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c algorithm 750, collected algorithms from acm.
c this work published in transactions on mathematical software,
c vol. 21, no. 4, december, 1995, p. 410--415.
c *************************************************************************
subroutine visitor(n,a,nstac,iroad,x,fstar)
integer a(n,n),x(nstac)
c real alpha,avson,spars
c integer active,err,i,inf,lb0,lbc,lopt,maxnd,n,nass,nexp,nprobq,
c + ordx,zeur,zstar
integer fstar(n),iroad(n),active,err,ordx,zeur,zstar
do 11,i=1,n
iroad(i)=0
11 continue
ordx = nstac
inf = 99999999
alpha = -1.
maxnd = -1
zeur = -1
do 1 i=1,nstac
x(i)=0.
1 continue
do 2 i=1,n
do 2 j=1,n
k=i+(j-1)*n
x(k)=a(i,j)
2 continue
c
call cdt(n,ordx,x,maxnd,inf,alpha,zeur,zstar,fstar,lb0,lbc,nexp,
+ nprobq,nass,active,lopt,spars,avson,err)
c
if (err.eq.0) go to 10
c wwfmt='('' solution not optimal '')'
10 continue
iroad(1)=1
do 20 i=2,n
iroad(i)=fstar(iroad(i-1))
20 continue
return
end
c *************************************************************************
subroutine cdt(n,ordx,x,maxnd,inf,alpha,zeur,zstar,fstar,lb0,lbc,
+ nexp,nprobq,nass,active,lopt,spars,avson,err)
c i/o parameter --------------------------------------------------------
c input parameters
c alpha = if .gt. 0 define the artificial upper bound ub=alpha*lb0
c if .le. 0 no effect
c n = number of nodes
c inf = very large number
c maxnd = if greater or equal to zero it is the maximum number of
c node to be explored on the branch decision-tree
c ordx = dimension of vector x
c zeur = if greater then zero it must define a valid upper bound
c the cost matrix is stored (column by column) in the first n**2 elem.
c of the vector x. the cost matrix it is not available in output.
c
c output value
c
c active = number of active problems in queue when the program stops
c (not yet and not currently examined)
c avson = average number of son nodes for the explored problems
c err = .gt. zero if an error occurred
c = -1 if no solution with value less than the artificial
c upper bound was found.
c in this case zstar is the artificial upper bound value
c fstar(i)= successor of node i in the current best solution
c lb0 = value of the optimal assignment at the root node
c lbc = value of the current best lb, i.e. value of the lower
c bound of the last problem extracted from the queue;
c lbc is a valid lower bound for the instance
c nass = number of assignment solved
c nexp = number of explored problems
c nprobq = number of problems stored in the queue
c spars = sparsity of the reduced matrix
c zeur = minimum value between the value of zeur in input and the
c value of the approximate solution computed using the
c patching heuristic
c zstar = value of the current best solution
c
c internal storage of vectors and matrices in vector x
c
c x(mm1) = the queue v(ordv) (dinamically allocated)
c x(mm2) = first element of matrices mv,mf (dinamically allocated)
c x(mm3) = vector ic(*)
c x(mm4) = vectors c(*) (cra(*) in ctcs,
c x(mm5) = ica(*) in ctcs)
c x(mm6) = iva(n)
c x(mm7) = ivb(n)
c x(mm8) = cr(n+1)
c x(mm9) = primal solution of the map associated with the current
c branched node f(n)
c x(mm10) = dual var. of the columns in the solution of ap, dualv(n)
c x(mm11) = primal solution of the map at the current node fc(n)
c x(mm12) = current dual variables of map, vd(n)
c x(mm13) = the heuristic solution of atsp
c x(mm14)..x(mm26) = temporary vectors
c
c fmvf = pointer to the first free element to store columns of
c matrices mv,mf
c x(fmvf) = pointer to the next free element for matrices mv,mf
c
c local variables
c .. scalar arguments ..
c real alpha,avson,spars
c integer active,err,inf,lb0,lbc,lopt,maxnd,n,nass,nexp,nprobq,ordx,
c + zeur,zstar
integer active,err,ordx,zeur,zstar
c .. array arguments ..
integer fstar(n),x(ordx)
c .. local scalars ..
c integer fmvf,i,ibig,idelta,ienlrg,imp,inf2,ipp,isalva,
c + isalvi,ivai,k,maxdix,maxica,mm1,mm10,mm11,mm12,mm13,mm14,
c + mm15,mm16,mm17,mm18,mm19,mm2,mm20,mm21,mm22,mm23,mm24,
c + mm25,mm26,mm3,mm4,mm5,mm6,mm7,mm8,mm9,nc,nci,ncodal,ngen,
c + nlsp,nnodi,nnodin,nodoim,nprob,nprobv,nq,offv,orcr,orcra,
c + ordsp,ordv,pnuovo,psalvo,punta,puntb,puntlv,r,sc1,totass,
c + ur,vimpa,vimpb,z,zarf,zc,zeri
integer fmvf,offv,orcr,orcra,ordsp,ordv,pnuovo,psalvo,punta,
+puntb,puntlv,r,sc1,totass,ur,vimpa,vimpb,z,zarf,zc,zeri
logical artif,eur
c subroutines ..
c actpro,agmhp,calcud,calcur,cercsb,clearq,contci,copyx,creams,ctcs,
c enlarg,enlini,errors,exque,genson,inque,inquer,karp
c ..
inf2 = float(inf)/2.
nq = n**2
err = 0
c
mm26 = ordx - n + 1
mm25 = mm26 - n
mm24 = mm25 - n
mm23 = mm24 - n
mm22 = mm23 - n
mm21 = mm22 - n
mm20 = mm21 - n
mm19 = mm20 - n
mm18 = mm19 - n
mm17 = mm18 - n
mm16 = mm17 - n
mm15 = mm16 - n
mm14 = mm15 - n
mm13 = mm14 - n
mm12 = mm13 - n
mm11 = mm12 - n
mm10 = mm11 - n
mm9 = mm10 - n
mm8 = mm9 - n - 1
mm7 = mm8 - n
mm6 = mm7 - n
if (mm6.gt.nq) then
mm5 = nq + (mm6-nq)*2./3.
mm4 = nq + (mm6-nq)/3.
mm3 = nq + 1
maxica = mm6 - mm5 + 1
orcra = mm5 - mm4 + 1
maxdix = mm4 - mm3 + 1
orcr = n + 1
c
nprob = 1
psalvo = 0
ncodal = 0
puntlv = 1
vimpa = 0
vimpb = 0
sc1 = -1
ngen = 1
fmvf = 0
nass = 1
c statistics information
nexp = 1
nprobq = 1
lopt = 1
avson = 0.
spars = 0.
active = 0
c solve the initial assignment problem
call ctcs(n,z,1,x(1),x(mm9),x(mm13),x(mm14),x(mm15),x(mm16),
+ x(mm17),x(mm4),x(mm5),x(mm8),x(mm3),x(mm26),x(mm10),
+ x(mm20),x(mm21),x(mm23),x(mm22),x(mm24),maxdix,
+ maxica,x(mm25),orcr,orcra,inf2,err)
if (err.eq.0) then
lb0 = z
lbc = z
c
c totass stores the value of the ap solution at the root node
totass = z
call contci(x(mm9),nci,n,x(mm14))
call enlini(n,x(mm8),x(mm3),x(mm9),x(mm26),x(mm10),
+ x(mm14),x(mm15),x(mm16),x(mm17),x(1),x(mm4),
+ x(mm5),nc,inf2)
zstar = totass
if (nc.eq.1) then
c optimal solution found at the root node by ctcs or enlini
call copyx(x(mm9),fstar,n)
zstar = totass
return
else
c
call copyx(x(mm9),x(mm13),n)
call karp(n,x(1),x(mm13),x(mm15),x(mm16),x(mm17),
+ x(mm18),totass,zstar,inf2)
c check if the heuristic solution is an optimum
if (zeur.gt.0 .and. zeur.lt.zstar) then
eur = .true.
zstar = zeur
else
eur = .false.
call copyx(x(mm13),fstar,n)
end if
if (zstar.eq.totass) return
avson = nc
artif = .false.
if (alpha.gt.0.0) then
if (totass*alpha.lt.zstar) then
zstar = totass*alpha
zarf = zstar
artif = .true.
end if
end if
c
idelta = zstar - lb0
call creams(idelta,x(1),n,x(mm8),x(1),mm6,mm4,mm3,
+ x(mm26),x(mm10),zeri,nlsp,err)
if (err.eq.0) then
ordsp = nlsp
spars = nlsp*100./float(n* (n-1))
ienlrg = 0
if (float(zeri).gt.2.5*float(n)) ienlrg = 1
c
c choose the subtour for the branch phase
c
call cercsb(x(mm9),x(mm14),x(mm10),n,nnodi,isalva,
+ inf2)
nnodin = nnodi
isalvi = isalva
c
c define the working arrays to store the queue:
c scalar and vectorial information
mm1 = 1
mm2 = mm3
ordv = mm2 - mm1
c
c mm1 = first element containing scalar informations
c mm2 = first position occupied by matrices mf,mv
c
c insert the root node in queue and initialize the vectors
c for the branching phase
puntlv = 1
call inquer(isalvi,nnodin,psalvo,ngen,n,totass,
+ vimpa,vimpb,x(mm9),x(mm10),puntlv,
+ punta,puntb,ordv,x(mm6),x(mm7),x(mm1),
+ pnuovo,nprob,sc1,offv,err)
if (err.eq.0) then
psalvo = pnuovo
active = 0
c ww(6,fmt='('' root node: zstar='',i10,'' lb0='',i10)') zstar,lb0
else
call errors(err,4)
return
end if
else
call errors(err,3)
return
end if
end if
else
call errors(err,2)
return
end if
else
call errors(err,1)
return
end if
c psalvo is the pointer to the actual problem
100 do 200 i = 1,nprob
c
c generate the descending nodes of psalvo
c
call genson(n,x(mm6),x(mm7),x(mm10),nprob,i,vimpa,vimpb,
+ nodoim,x(mm8),ordsp,x(mm3),x(mm4),inf2)
call copyx(x(mm9),x(mm11),n)
call copyx(x(mm10),x(mm12),n)
zc = totass
ivai = x(mm6+i-1)
c
c solve the new map problem
c
call calcur(n,x(mm8),ordsp,x(mm3),x(mm4),x(mm11),ivai,x(mm12),
+ ur,inf2)
call agmhp(n,ivai,x(mm8),ordsp,x(mm3),x(mm4),x(mm11),x(mm12),
+ ur,zc,x(mm14),x(mm15),x(mm16),x(mm17),x(mm18),
+ zstar,imp,inf2)
if (nass.eq.maxnd) then
call errors(err,-1)
return
else if (imp.ne.1) then
if (zc.lt.zstar) then
nass = nass + 1
c
c count the subtours
c
call contci(x(mm11),nc,n,x(mm15))
if (ienlrg.ne.0) then
call calcud(n,ordsp,x(mm3),x(mm8),x(mm4),x(mm12),
+ x(mm11),x(mm14))
call enlarg(n,x(mm8),ordsp,x(mm3),x(mm4),x(mm11),
+ x(mm14),x(mm12),x(mm15),x(mm16),
+ x(mm17),x(mm18),nc,inf2)
end if
if (nc.eq.1) then
c
c found a new feasible solution
c
zstar = zc
c
c clear the queue of the active problems
c
call clearq(zstar,puntb,fmvf,ordx,x(1),ncodal,
+ ordv,x(mm1),active)
call copyx(x(mm11),fstar,n)
c
c wwfmt='('' n.exp='',i8,'' n.prob.q='',i8,'' n.ass='',i8, '' active='
c ',i8,'' av.son='',f8.2)' nexp,nprobq,nass,active,avson
c
c wwfmt='('' zstar='',i10,'' lbc='',i10,'' nprobq='',i8,'' active='',i8)'
c zstar,lbc,nprobq,active
c statistics
lopt = 2
ipp = psalvo
105 ipp = x(mm1+ipp+2)
lopt = lopt + 1
if (ipp.gt.1) go to 105
c
c if the cost zc of the solution of the current problem p is ugual to
c the lb of the problem father of p, say pf, no other son of pf
c have to be generated
c
ibig = x(mm1+psalvo+1)
if (zc.le.ibig) then
nprobv = i
go to 300
end if
else
call cercsb(x(mm11),x(mm14),x(mm12),n,nnodi,
+ isalva,inf2)
ngen = i
c
c insert the problem in queue
c
if (fmvf.ne.0) then
k = fmvf
fmvf = x(fmvf)
else
mm2 = mm2 - 2*n
k = mm2
end if
ordv = mm2 - mm1
if (mm1+puntlv+offv+nnodi.lt.mm2) then
call inque(isalva,nnodi,sc1,psalvo,ngen,
+ x(mm12),x(mm11),n,zc,x(k),x(k+n),k,
+ vimpa,vimpb,ordv,x(mm1),ncodal,
+ puntlv,punta,puntb,offv,inf2)
nprobq = nprobq + 1
active = active + 1
r = int(nprobq/1000.)*1000
if (r.eq.nprobq) continue
c wwfmt='('' zstar='',i10,'' lbc='',i10,'' nprobq='',i8, '' active='
c ',i8)' zstar,lbc,nprobq,active
else
call errors(err,5)
return
end if
end if
end if
end if
c
200 continue
c
nprobv = nprob
300 if (ncodal.ne.0) then
c
c extract the problem with lowest bound
call exque(ordv,x(mm1),ordx,x(1),fmvf,x(mm10),x(mm9),
+ pnuovo,ngen,punta,nprob,x(mm6),x(mm7),ncodal,n,sc1,
+ totass,offv)
avson = (avson*nexp+nprob)/float(nexp+1)
nexp = nexp + 1
active = active - 1
if (lbc.lt.totass) then
lbc = totass
c wwfmt='('' zstar='',i10,'' lbc='',i10,'' nprobq='',
c i8,'' active='',i8)' zstar,lbc,nprobq,active
end if
lbc = totass
c
c update sets of included and excluded arcs
c
call actpro(psalvo,n,nprobv,pnuovo,ngen,vimpa,vimpb,ordv,
+ x(mm1),x(mm8),ordsp,x(mm3),x(mm4),offv,inf2)
psalvo = pnuovo
go to 100
end if
c
c
c the tree search is terminated
c
if (artif .and. zstar.eq.zarf) call errors(err,6)
if (eur) then
c wwfmt='('' zeur is the optimal solution value'')'
c wwfmt='('' fstar() do not contain the optimal tour'')'
end if
return
end
c *************************************************************************
subroutine contci(f,nc,n,flag)
integer f(n),flag(n)
c .. local scalars i,k
nc = 0
do 100 i = 1,n
flag(i) = 0
100 continue
do 200 i = 1,n
if (flag(i).le.0) then
k = i
nc = nc + 1
120 flag(k) = 1
k = f(k)
if (k.ne.i) go to 120
end if
200 continue
return
end
c *************************************************************************
subroutine loadfv(f,v,n,vroot,froot)
integer f(n),froot(n),v(n),vroot(n)
do 100 i = 1,n
froot(i) = f(i)
vroot(i) = v(i)
100 continue
return
end
c *************************************************************************
subroutine backfv(vroot,froot,n,vd,f)
integer f(n),froot(n),vd(n),vroot(n)
do 100 i = 1,n
f(i) = froot(i)
vd(i) = vroot(i)
100 continue
return
end
c *************************************************************************
subroutine inquer(isalva,nnodi,psalvo,ngen,n,totass,vimpa,vimpb,f,
+ vd,puntlv,punta,puntb,ordv,iva,ivb,v,p2p,nprob,
+ sc1,offv,ierr)
integer offv,ordv,p2p,psalvo,punta,
+ puntb,puntlv,sc1,totass,vimpa,vimpb
integer f(n),iva(n),ivb(n),v(ordv),vd(n)
integer i,i32,ksalva,nm1,nodob,nodopa,plvp
c ..
i32 = 32000
offv = 7
if (puntlv+offv+nnodi.le.ordv) then
nm1 = ngen - 1
sc1 = sc1 + (nm1)*10 + 1
v(puntlv+2) = totass
v(puntlv+3) = psalvo
v(puntlv+4) = ngen*i32 + nnodi
v(puntlv+5) = 1
v(puntlv+6) = vimpa*i32 + vimpb
v(puntlv+7) = sc1
plvp = puntlv + offv
i = 1
ksalva = isalva
nodopa = isalva
50 nodob = f(nodopa)
v(plvp+i) = nodopa*i32 + nodob
iva(i) = nodopa
ivb(i) = nodob
i = i + 1
nodopa = nodob
if (nodopa.ne.ksalva) go to 50
punta = puntlv
puntb = puntlv + 1
puntlv = puntlv + offv + nnodi + 1
p2p = punta
nprob = nnodi
sc1 = 0
return
end if
ierr = 2
return
end
c *************************************************************************
subroutine copyx(f,fc,n)
integer f(n),fc(n)
do 100 i = 1,n
fc(i) = f(i)
100 continue
return
end
c *************************************************************************
subroutine ctcs(n,z,ks,a,f,r,vs,fs,uold,sur,cra,ica,cr,ic,u,v,fb,
+ pi,c,dm,ll,maxdix,maxica,rx,orcr,orcra,inf,ierr)
c solution of the linear min-sum assignment problem through the
c algorithm presented in the paper:
c g. carpaneto and p. toth, "primal-dual algorithms for the
c assignment problem", discrete applied mathematics 18, 137-153,1987
integer ierr,inf,ks,maxdix,maxica,n,orcr,orcra,z
integer a(n,n),c(n),cr(orcr),cra(orcra),dm(n),f(n),fb(n),fs(n),
+ ic(maxdix),ica(maxica),ll(n),pi(n),r(n),rx(n),sur(n),u(n),
+ uold(n),v(n),vs(n)
c .. local scalars ..
integer i,iflag,j,k,kfeas,kopt,krip,kth,m,m1,max,maxdim,
+ maxfea,maxrip,nfeas,nkk,nur,zz
c subroutines apmmix,asmixm,feaso,indus3,opto,setupo
c .. data statements ..
data maxrip/5/,maxfea/1/
c ..
maxdim = maxdix - n
krip = 0
nfeas = 0
do 100 i = 1,n
f(i) = 0
fb(i) = 0
cra(i) = 0
a(i,i) = inf
100 continue
nkk = 0
cr(1) = -1
m1 = -1
call indus3(n,a,f,m,u,v,fb,pi,max,inf)
z = 0
zz = 0
do 200 i = 1,n
z = z + v(i)
zz = zz + u(i)
200 continue
if (m.ne.n) then
if (ks.eq.1) then
c
if (float(m)/float(n).gt.0.6) then
ierr = 0
call setupo(n,a,u,v,m,ic,cr,kth,maxdim,iflag,inf)
if (iflag.ne.1) then
if (cr(1).ge.0) then
do 202 i = 1,n
vs(i) = v(i)
fs(i) = f(i)
uold(i) = u(i)
202 continue
204 call asmixm(n,a,ic,cr,f,fb,u,v,z,pi,r,c,cra,
+ ica,sur,nur,dm,ll,rx,inf)
if (ierr.ne.0) return
if (nur.gt.0) then
if (nfeas.ne.maxfea) then
nfeas = nfeas + 1
call feaso(n,a,kfeas,kth,uold,vs,
+ nfeas,cra,ica,nkk,sur,nur,
+ maxica)
if (kfeas.eq.1) go to 204
end if
do 206 j = 1,n
fb(j) = 0
206 continue
do 208 i = 1,n
j = fs(i)
if (j.gt.0) fb(j) = i
u(i) = uold(i)
v(i) = vs(i)
f(i) = fs(i)
208 continue
else
call opto(n,a,u,v,kopt,cra,ica,nkk,f,fb,
+ uold,maxica,ierr)
if (ierr.eq.0) then
if (kopt.eq.1) return
if (kopt.ne.m1) then
if (krip.ne.maxrip) then
krip = krip + 1
go to 204
end if
end if
end if
end if
end if
end if
end if
end if
call apmmix(n,a,f,z,fb,u,v,pi,r,c,dm,ll,inf)
return
end if
z = 0
do 300 k = 1,n
z = z + u(k) + v(k)
300 continue
return
end
c *************************************************************************
subroutine opto(n,a,u,v,kopt,cra,ica,nkk,f,fb,uold,maxica,ierr)
c check the feasibility of the dual solution , and hence the
c optimality of the primal solution found by the sparse matrix
c procedure.
integer ierr,kopt,maxica,n,nkk
integer a(n,n),cra(*),f(n),fb(n),ica(maxica),u(n),uold(n),v(n)
integer i,ia,j,jc,k,kk,kkn,min,neg
c ..
kopt = 0
kk = nkk
neg = 0
do 100 i = 1,n
if (u(i).ne.uold(i)) then
min = 0
k = i
do 20 j = 1,n
ia = a(i,j) - u(i) - v(j)
if (ia.lt.0) then
if (ia.lt.0) neg = neg + 1
kk = kk + 1
if (kk.le.maxica) then
kkn = kk + n
2 if (cra(k).eq.0) then
cra(k) = kkn
cra(kkn) = 0
ica(kk) = j
else
k = cra(k)
go to 2
end if
end if
if (ia.lt.min) min = ia
end if
20 continue
u(i) = u(i) + min
uold(i) = u(i)
if (min.ne.0) then
jc = f(i)
f(i) = 0
fb(jc) = 0
kopt = kopt - 1
end if
end if
100 continue
nkk = kk
if (nkk.le.maxica) then
if (kopt.lt.0) then
kopt = 0
return
else
kopt = 1
return
end if
end if
ierr = 1
kopt = -1
return
end
c *************************************************************************
subroutine asmixm(n,a,ic,cr,f,fb,dualu,dualv,z,pre,uv,sr,cra,ica,
+ sur,nur,dm,ll,r,inf)
c shortest augmenting path and hungarian method :
c version for sparse matrices
integer inf,n,nur,z
integer a(n,n),cr(*),cra(*),dm(n),dualu(n),dualv(n),f(n),fb(n),
+ ic(*),ica(*),ll(n),pre(n),r(n),sr(n),sur(n),uv(n)
integer aa,d,i,i1,i2,i2p1,ii,iix,ik,ind,index,indexv,iv,j,j1,k,k1,
+ k2,kdir,kmn,knuv,li,nuv,u,w
c ..
nur = 0
do 500 u = 1,n
if (f(u).gt.0) go to 500
do 50 i = 1,n
ll(i) = 0
dm(i) = inf
50 continue
nuv = n
j1 = 0
knuv = 0
k1 = cr(u)
k2 = cr(u+1) - 1
do 100 ik = k1,k2
i = ic(ik)
dm(i) = a(u,i) - dualu(u) - dualv(i)
pre(i) = u
j1 = j1 + 1
sr(j1) = i
100 continue
k = u
150 if (cra(k).eq.0) then
r(1) = u
else
k = cra(k)
kmn = k - n
i = ica(kmn)
dm(i) = a(u,i) - dualu(u) - dualv(i)
pre(i) = u
j1 = j1 + 1
sr(j1) = i
go to 150
end if
200 d = inf
index = 0
i2 = 1
i1 = i2
c
if (nuv.lt.j1) then
kdir = 1
if (knuv.ne.1) then
iv = 0
do 210 i = 1,n
if (ll(i).ne.1) then
iv = iv + 1
uv(iv) = i
end if
210 continue
knuv = 1
end if
do 220 iv = 1,nuv
i = uv(iv)
if (dm(i).le.d) then
if (dm(i).ne.d) then
index = 0
i2 = i1
end if
d = dm(i)
if (fb(i).le.0) then
index = i
if (d.eq.0) go to 300
end if
i2 = i2 + 1
r(i2) = iv
end if
220 continue
else
kdir = 0
c
do 240 li = 1,j1
i = sr(li)
if (dm(i).le.d) then
if (ll(i).ne.1) then
if (dm(i).ne.d) then
index = 0
i2 = i1
end if
d = dm(i)
if (fb(i).le.0) then
index = i
if (d.eq.0) go to 300
end if
i2 = i2 + 1
r(i2) = i
end if
end if
240 continue
end if
if (d.eq.inf) then
c
nur = nur + 1
sur(nur) = u
go to 500
else if (index.le.0) then
i1 = i1 + 1
i2p1 = i2 + i1
c
do 280 iix = i1,i2
ii = i2p1 - iix
if (kdir.eq.1) then
indexv = r(ii)
index = uv(indexv)
uv(indexv) = uv(nuv)
else
index = r(ii)
end if
ll(index) = 1
nuv = nuv - 1
w = fb(index)
k1 = cr(w)
k2 = cr(w+1) - 1
c
do 250 ik = k1,k2
i = ic(ik)
if (ll(i).ne.1) then
aa = d + a(w,i) - dualu(w) - dualv(i)
if (dm(i).gt.aa) then
if (dm(i).ge.inf) then
j1 = j1 + 1
sr(j1) = i
end if
dm(i) = aa
pre(i) = w
end if
end if
250 continue
k = w
260 if (cra(k).ne.0) then
k = cra(k)
kmn = k - n
i = ica(kmn)
if (ll(i).ne.1) then
aa = d + a(w,i) - dualu(w) - dualv(i)
if (dm(i).gt.aa) then
if (dm(i).ge.inf) then
j1 = j1 + 1
sr(j1) = i
end if
dm(i) = aa
pre(i) = w
end if
end if
go to 260
end if
280 continue
go to 200
end if
c
300 do 350 j = 1,n
if (dm(j).lt.d) then
dualv(j) = dualv(j) + dm(j) - d
i = fb(j)
dualu(i) = dualu(i) - dm(j) + d
end if
350 continue
dualu(u) = dualu(u) + d
c
400 w = pre(index)
fb(index) = w
ind = f(w)
f(w) = index
if (w.ne.u) then
index = ind
go to 400
end if
500 continue
if (nur.le.0) then
c
z = 0
do 550 i = 1,n
z = z + dualu(i) + dualv(i)
550 continue
return
end if
z = -1
return
end
c *************************************************************************
subroutine feaso(n,a,kfeas,kth,us,vs,nfeas,cra,ica,nkk,sur,nur,
+ maxica)
c insert new entries in rows sur (l) (l = 1,nur) in order to find
c a feasible assignment in the sparse cost matrix.
integer kfeas,kth,maxica,n,nfeas,nkk,nur
integer a(n,n),cra(*),ica(maxica),sur(n),us(n),vs(n)
integer coef,i,ia,ii,iukth,iukthn,j,k,kk,kkn,kth0,kthn
data coef/3/
c ..
kk = nkk
kth0 = kth
if (kth0.eq.0) kth0 = 1
do 200 ii = 1,nur
i = sur(ii)
cc = 0.
50 cc = cc + float(coef*nfeas)
kthn = cc*float(kth0)
iukthn = kthn + us(i)
iukth = kth + us(i)
k = i
do 100 j = 1,n
ia = a(i,j) - vs(j)
if (ia.le.iukthn) then
if (ia.gt.iukth) then
kk = kk + 1
if (kk.gt.maxica) go to 300
kkn = kk + n
cra(k) = kkn
cra(kkn) = 0
ica(kk) = j
end if
end if
100 continue
if (cra(i).eq.0) go to 50
200 continue
nkk = kk
kfeas = 1
return
300 kfeas = -1
return
end
c
c *************************************************************************
subroutine indus3(n,a,f,m,u,v,fb,fu,max,inf)
c search for an initial dual solution ( u(i),v(j) ) and an initial
c partial primal solution ( f(i),fb(j) ) of the ap problem
integer inf,m,max,n
integer a(n,n),f(n),fb(n),fu(n),u(n),v(n)
integer i,ia,ii,imin,j,j1,jj,jmin,l,maxl,min
c ..
m = 0
do 100 j = 1,n
u(j) = 0
min = inf
do 50 i = 1,n
ia = a(i,j)
if (ia.le.min) then
if (ia.ge.min) then
if (f(i).ne.0) go to 50
end if
min = ia
imin = i
end if
50 continue
v(j) = min
if (f(imin).eq.0) then
m = m + 1
fb(j) = imin
f(imin) = j
fu(imin) = j + 1
end if
100 continue
max = 0
do 400 i = 1,n
if (f(i).ne.0) go to 400
min = inf
maxl = 0
do 150 j = 1,n
l = a(i,j) - v(j)
if (l.gt.max) maxl = l
if (l.le.min) then
if (l.ge.min) then
if (fb(j).ne.0) go to 150
if (fb(jmin).eq.0) go to 150
end if
min = l
jmin = j
end if
150 continue
u(i) = min
if (maxl-min.gt.max) max = maxl - min
j = jmin
if (fb(j).eq.0) go to 300
do 200 j = jmin,n
if (a(i,j)-v(j).le.min) then
ii = fb(j)
j1 = fu(ii)
if (j1.le.n) then
do 155 jj = j1,n
if (fb(jj).le.0) then
if (a(ii,jj)-u(ii).eq.v(jj)) go to 250
end if
155 continue
fu(ii) = n + 1
end if
end if
200 continue
go to 400
250 f(ii) = jj
fb(jj) = ii
fu(ii) = jj + 1
300 m = m + 1
fb(j) = i
f(i) = j
fu(i) = j + 1
400 continue
return
end
c *************************************************************************
subroutine apmmix(n,a,f,z,fb,dualu,dualv,pre,uv,r,dm,dp,inf)
c shortest augmenting path and hungarian method for complete matrices
integer inf,n,z
integer a(n,n),dm(n),dp(n),dualu(n),dualv(n),f(n),fb(n),pre(n),
+ r(n),uv(n)
integer d,i,i1,i2,i2p1,ii,iix,ind,index,indexv,iv,nuv,u,vgl,w
c ..
do 400 u = 1,n
if (f(u).gt.0) go to 400
do 50 i = 1,n
pre(i) = u
uv(i) = i
dp(i) = inf
dm(i) = a(u,i) - dualu(u) - dualv(i)
50 continue
nuv = n
i2 = 0
r(1) = u
dp(u) = 0
100 d = inf
index = 0
i1 = i2
do 150 iv = 1,nuv
i = uv(iv)
if (dm(i).le.d) then
if (dm(i).ne.d) then
index = 0
i2 = i1
end if
d = dm(i)
if (fb(i).le.0) then
if (d.eq.0) go to 250
index = i
end if
i2 = i2 + 1
r(i2) = iv
end if
150 continue
if (index.gt.0) go to 300
i1 = i1 + 1
i2p1 = i2 + i1
do 200 iix = i1,i2
ii = i2p1 - iix
indexv = r(ii)
index = uv(indexv)
uv(indexv) = uv(nuv)
nuv = nuv - 1
w = fb(index)
dp(w) = d
do 160 iv = 1,nuv
i = uv(iv)
vgl = d + a(w,i) - dualu(w) - dualv(i)
if (dm(i).gt.vgl) then
dm(i) = vgl
pre(i) = w
end if
160 continue
200 continue
go to 100
c
250 index = i
300 w = pre(index)
fb(index) = w
ind = f(w)
f(w) = index
if (w.eq.u) then
c
do 320 i = 1,n
if (dp(i).ne.inf) dualu(i) = dualu(i) + d - dp(i)
if (dm(i).lt.d) dualv(i) = dualv(i) + dm(i) - d
320 continue
else
index = ind
go to 300
end if
400 continue
c
z = 0
do 500 i = 1,n
z = z + dualu(i) + dualv(i)
500 continue
return
end
c *************************************************************************
subroutine agmhp(n,r,cr,ordsp,ic,c,f,v,ur,z,p,q,hp,dm,fb,zs,imp,
+ inf)
c agmheap search for an augmenting path starting at row r, using
c johnson algorithm, implemented via a heap-queue
integer imp,inf,n,ordsp,r,ur,z,zs
integer c(ordsp),cr(n+1),dm(n),f(n),fb(n),hp(n),ic(ordsp),p(n),
+ q(n),v(n)
integer delta,dj,dp1,du,fb1,fbj,h,hp1,hp2,hp3,i,ifin,ind,iniz,j,k,
+ k2,nhp,w
c ..
delta = zs - z + ur
imp = 0
f(r) = 0
do 100 i = 1,n
fb(i) = 0
100 continue
do 200 i = 1,n
q(i) = 0
p(i) = 0
j = f(i)
if (j.gt.0) fb(j) = i
dm(i) = inf
200 continue
nhp = 0
p(r) = 0
du = 0
w = r
iniz = cr(w)
ifin = cr(w+1) - 1
if (iniz.le.ifin) go to 500
go to 700
300 w = fb(i)
c
iniz = cr(w)
ifin = cr(w+1) - 1
if (iniz.gt.ifin) go to 700
c
do 400 h = iniz,ifin
j = ic(h)
if (j.eq.i) then
du = du - c(h)
go to 500
end if
400 continue
c
500 do 600 h = iniz,ifin
j = ic(h)
dj = du + c(h) - v(j)
if (w.ne.r) dj = dj + v(i)
if (dm(j).gt.dj) then
dm(j) = dj
p(j) = w
fbj = fb(j)
if (q(j).eq.0) then
c
nhp = nhp + 1
q(j) = nhp
end if
c
k = q(j)
520 k2 = k/2
if (k2.gt.0) then
hp2 = hp(k2)
if (dj.le.dm(hp2)) then
if (dj.ge.dm(hp2)) then
if (fbj.ne.0 .or. fb(hp2).eq.0) go to 540
end if
hp(k) = hp2
q(hp2) = k
k = k2
go to 520
end if
end if
540 hp(k) = j
q(j) = k
end if
600 continue
c
700 i = hp(1)
du = dm(i)
if (du.gt.delta) go to 1000
if (fb(i).eq.0) then
750 w = p(i)
fb(i) = w
ind = f(w)
f(w) = i
if (w.ne.r) then
c
i = ind
go to 750
else
c
do 760 i = 1,n
if (dm(i).lt.du) v(i) = v(i) + dm(i) - du
760 continue
z = z + du - ur
return
end if
else
q(i) = 0
nhp = nhp - 1
c
if (nhp.lt.0) then
c
imp = 1
return
else if (nhp.eq.0) then
go to 300
else
c
hp1 = hp(nhp+1)
dp1 = dm(hp1)
fb1 = fb(hp1)
k = 1
end if
end if
800 k2 = k*2
if (k2.lt.nhp) then
hp2 = hp(k2)
hp3 = hp(k2+1)
if (dm(hp2).ge.dm(hp3)) then
if (dm(hp2).ne.dm(hp3) .or. fb(hp2).ne.0) then
hp2 = hp3
k2 = k2 + 1
end if
end if
else if (k2.ne.nhp) then
go to 900
end if
hp2 = hp(k2)
if (dp1.le.dm(hp2)) then
if (dp1.lt.dm(hp2)) go to 900
if (fb1.eq.0 .or. fb(hp2).ne.0) go to 900
end if
hp(k) = hp2
q(hp2) = k
k = k2
go to 800
900 hp(k) = hp1
q(hp1) = k
go to 300
1000 z = inf
return
end
c *************************************************************************
subroutine enlini(n,cr,ic,f,u,v,s,p,flag,cont,a,cra,ica,nc,inf)
c try to connect subtours of the solution of the assignment at
c the root node, with zero reduced cost arcs
integer inf,n,nc
integer a(n,n),cont(n),cr(n+1),cra(*),f(n),flag(n),ic(*),ica(*),
+ p(n),s(n),u(n),v(n)
integer card,corid,flaga,fsalva,i,iall,ii,ij,ijn,ik,iolds,ip1,ip2,
+ istart,j,k,kc,kk,l,lm,lm2,m1,nc2,ninf2,pj,si,ult
c ..
ninf2 = -float(inf)/2.
m1 = -1
kc = 0
iall = 0
do 100 i = 1,n
j = f(i)
s(i) = j
p(j) = i
flag(i) = 0
cont(i) = 0
100 continue
do 200 istart = 1,n
if (flag(istart).ne.0) go to 200
kc = kc + 1
k = istart
card = 0
150 flag(k) = -kc
k = s(k)
if (v(k).gt.ninf2) card = card + 1
if (k.ne.istart) go to 150
if (card.ne.n) then
j = 2*kc
do 160 i = 1,j,2
if (cont(i).le.card) then
if (cont(i).ne.0) then
ip2 = i + 2
do 152 l = ip2,j
lm = j + ip2 - l
lm2 = lm - 2
cont(lm) = cont(lm2)
152 continue
end if
cont(i) = card
ip1 = i + 1
cont(ip1) = istart
go to 200
end if
160 continue
else
nc = 1
return
end if
200 continue
nc = kc
nc2 = nc*2
do 600 ii = 2,nc2,2
istart = cont(ii)
if (flag(istart).gt.0) go to 600
fsalva = iabs(flag(istart))
flaga = 0
i = istart
250 j = f(i)
if (v(j).le.ninf2) go to 550
if (cr(1).ne.m1) then
c
ip1 = i + 1
ult = cr(ip1) - 1
ik = i
k = cr(i)
j = ic(k)
go to 400
end if
300 k = 1
ult = n
350 j = k
c
400 if (i.eq.j) go to 500
450 corid = a(i,j) - u(i) - v(j)
if (corid.le.0) then
if (iabs(flag(j)).ne.fsalva) then
if (flaga.ne.0 .or. flag(j).le.0) then
pj = p(j)
if (v(j).ge.ninf2) then
si = f(i)
if (a(pj,si)-u(pj).le.v(si)) then
c
flaga = 1
iall = iall + 1
nc = nc - 1
si = f(i)
pj = p(j)
f(i) = j
p(j) = i
f(pj) = si
p(si) = pj
iolds = s(i)
kk = j
if (flag(j).gt.0) s(i) = pj
if (flag(j).le.0) s(i) = j
s(pj) = iolds
452 flag(kk) = fsalva
kk = f(kk)
if (kk.ne.si) go to 452
if (cr(1).eq.m1) go to 300
k = cr(i)
j = ic(k)
go to 400
end if
end if
end if
end if
end if
500 k = k + 1
if (cr(1).ne.m1) then
if (k.le.ult) then
j = ic(k)
go to 400
else if (cra(ik).ne.0) then
ij = cra(ik)
ik = ij
ijn = ij - n
j = ica(ijn)
go to 450
end if
else if (k.le.ult) then
go to 350
end if
550 flag(i) = iabs(flag(i))
i = s(i)
if (i.ne.istart) go to 250
600 continue
return
end
c *************************************************************************
subroutine enlarg(n,cr,ordsp,ic,c,f,u,vd,s,p,flag,cont,nc,inf)
c try to connect subtours of the current modified assignment
c problem, with zero reduced cost arcs
integer inf,n,nc,ordsp
integer c(ordsp),cont(n),cr(n+1),f(n),flag(n),ic(ordsp),p(n),s(n),
+ u(n),vd(n)
integer card,corid,flaga,fsalva,i,iall,ii,incr,iolds,ip1,ip2,ipj,
+ istart,j,jj,k,kc,kk,l,lm,lm2,m1,meta,minf2,nc2,pj,si,ult,
+ ult2
c ..
minf2 = -float(inf)/2.
kc = 0
m1 = -1
iall = 0
do 100 i = 1,n
j = f(i)
s(i) = j
p(j) = i
flag(i) = 0
cont(i) = 0
100 continue
do 200 istart = 1,n
if (flag(istart).ne.0) go to 200
kc = kc + 1
k = istart
card = 0
150 flag(k) = -kc
k = s(k)
if (vd(k).gt.minf2) card = card + 1
if (k.ne.istart) go to 150
if (card.ne.n) then
j = 2*kc
do 160 i = 1,j,2
if (cont(i).le.card) then
if (cont(i).ne.0) then
ip2 = i + 2
do 152 l = ip2,j
lm = j + ip2 - l
lm2 = lm - 2
cont(lm) = cont(lm2)
152 continue
end if
cont(i) = card
ip1 = i + 1
cont(ip1) = istart
go to 200
end if
160 continue
else
nc = 1
return
end if
200 continue
nc = kc
nc2 = nc*2
do 400 ii = 2,nc2,2
istart = cont(ii)
if (flag(istart).gt.0) go to 400
fsalva = iabs(flag(istart))
flaga = 0
i = istart
250 j = f(i)
if (vd(j).le.minf2) go to 350
ip1 = i + 1
ult = cr(ip1) - 1
k = cr(i)
300 j = ic(k)
corid = c(k) - u(i) - vd(j)
if (corid.le.0) then
if (iabs(flag(j)).ne.fsalva) then
if (flaga.ne.0 .or. flag(j).le.0) then
pj = p(j)
if (vd(j).ge.minf2) then
si = f(i)
ipj = pj + 1
ult2 = cr(ipj) - 1
kk = cr(pj)
meta = float(ic(ult2)-ic(kk))/2.
if (si.gt.meta) then
ipj = pj + 1
kk = cr(ipj) - 1
ult2 = cr(pj)
incr = -1
else
incr = 1
end if
302 jj = ic(kk)
if (jj.eq.si) then
if (c(kk)-u(pj).le.vd(si)) then
flaga = 1
iall = iall + 1
nc = nc - 1
si = f(i)
pj = p(j)
f(i) = j
p(j) = i
f(pj) = si
p(si) = pj
iolds = s(i)
kk = j
if (flag(j).gt.0) s(i) = pj
if (flag(j).le.0) s(i) = j
s(pj) = iolds
304 flag(kk) = fsalva
kk = f(kk)
if (kk.ne.si) go to 304
k = cr(i)
go to 300
end if
else if ((jj.le.si) .or. (incr.ne.1)) then
if ((jj.ge.si) .or. (incr.ne.m1)) then
if (kk.ne.ult2) then
kk = kk + incr
go to 302
end if
end if
end if
end if
end if
end if
end if
k = k + 1
if (k.le.ult) go to 300
350 flag(i) = iabs(flag(i))
i = s(i)
if (i.ne.istart) go to 250
400 continue
return
end
c *************************************************************************
subroutine karp(n,a,f,p,flag,na,nd,totass,ub,inf)
c heuristic solution of the atsp using the patching algorithm
integer inf,n,totass,ub
integer a(n,n),f(n),flag(n),na(n),nd(n),p(n)
integer cont,costo,fsalva,i,im1,istart,j,k,kk,kp,min,mn1,mn2,mn3,
+ nc,nim1,pmn1,rr,s2,temp,z
c ..
do 100 i = 1,n
flag(i) = 0
100 continue
cont = 0
nc = 0
do 200 i = 1,n
if (flag(i).le.0) then
k = i
nc = nc + 1
120 flag(k) = nc
cont = cont + 1
kk = f(k)
p(kk) = k
k = kk
if (k.ne.i) go to 120
na(nc) = cont
nd(nc) = i
cont = 0
end if
200 continue
if (nc.eq.1) return
c
c sort the subtours
c
ub = totass
z = nc
300 min = n
do 400 i = 1,z
if (na(i).lt.min) then
min = na(i)
k = i
end if
400 continue
na(k) = na(z)
na(z) = min
temp = nd(k)
nd(k) = nd(z)
nd(z) = temp
z = z - 1
if (z.ge.2) go to 300
rr = 2
500 istart = nd(rr)
i = istart
im1 = rr - 1
nim1 = nd(im1)
fsalva = flag(nim1)
mn2 = inf
600 do 700 k = 1,n
if (flag(k).eq.fsalva) then
j = f(i)
kp = p(k)
costo = a(i,k) + a(kp,j) - a(kp,k) - a(i,j)
if (costo.lt.mn2) then
mn1 = k
mn2 = costo
mn3 = i
if (costo.eq.0) go to 800
end if
end if
700 continue
i = f(i)
if (i.ne.istart) go to 600
800 s2 = f(mn3)
pmn1 = p(mn1)
f(mn3) = mn1
p(mn1) = mn3
f(pmn1) = s2
p(s2) = pmn1
ub = ub + mn2
c
c update flag
c
fsalva = flag(istart)
kk = mn1
900 flag(kk) = fsalva
kk = f(kk)
if (kk.ne.s2) go to 900
nc = nc - 1
if (nc.eq.1) return
rr = rr + 1
go to 500
end
c *************************************************************************
subroutine cercsb(f,flag,vd,n,cmin,isalva,inf)
c chose the subtour for the branch phase
integer cmin,inf,isalva,n
integer f(n),flag(n),vd(n)
integer card,i,istart,k,kc,minf2
c ..
minf2 = -float(inf)/2.
cmin = inf
kc = 0
do 100 i = 1,n
flag(i) = 0
100 continue
do 200 istart = 1,n
if (flag(istart).ne.0) go to 200
kc = kc + 1
card = 0
k = istart
150 flag(k) = kc
k = f(k)
if (vd(k).ge.minf2) card = card + 1
if (k.ne.istart) go to 150
if (card.lt.cmin) then
cmin = card
isalva = istart
end if
200 continue
return
end
c *************************************************************************
subroutine clearq(zstar,puntb,fmvf,ordx,x,ncodal,ordv,v,active)
c make not active nodes with lb greater than zstar
integer active,fmvf,ncodal,ordv,ordx,puntb,zstar
integer v(ordv),x(ordx)
integer j
c ..
100 if (ncodal.eq.0) return
if (v(puntb+1).lt.zstar) return
j = v(puntb+4)
x(j) = fmvf
fmvf = j
puntb = v(puntb)
ncodal = ncodal - 1
active = active - 1
go to 100
end
c *************************************************************************
subroutine inque(isalva,nnodi,sc1,psalvo,ngen,vd,f,n,totass,mv,mf,
+ mm2,vimpa,vimpb,ordv,v,ncodal,puntlv,punta,puntb,
+ ioffv,inf)
c psalvo = pointer to the father
c nnodi = number of nodes in the subtour
integer inf,ioffv,isalva,mm2,n,ncodal,ngen,nnodi,ordv,psalvo,
+ punta,puntb,puntlv,sc1,totass,vimpa,vimpb
integer f(n),mf(n),mv(n),v(ordv),vd(n)
integer i,i32,ksalva,lb,lb1,minf2,nm1,
+ nodob,nodopa,pcorr,plvp,puntol,sc2,sc3
c ..
i32 = 32000
c
c insert the problem in the queue in this order
c (from the third location):
c lb,
c psalvo,
c the generation number (among the sons of the same father) and the
c number of not imposed arcs in the subtour choosen for the branch
c phase (packed),
c the pointer to the column of matrices mv,mf,
c the additional arc to exclude from solution
c (start node + end node, packed),
c the score sc1 ((n.of excluded arcs)*10+number of imposed arcs),
c the not imposed arcs in the subtour choosen for the
c branch phase (packed)
c
minf2 = -float(inf)/2.
nm1 = ngen - 1
sc1 = sc1 + (nm1)*10 + 1
v(puntlv+2) = totass
v(puntlv+3) = psalvo
v(puntlv+4) = ngen*i32 + nnodi
v(puntlv+5) = mm2
do 100 i = 1,n
mv(i) = vd(i)
mf(i) = f(i)
100 continue
v(puntlv+6) = vimpa*i32 + vimpb
v(puntlv+7) = sc1
plvp = puntlv + ioffv
i = 1
ksalva = isalva
nodopa = isalva
200 nodob = f(nodopa)
if (vd(nodob).ge.minf2) then
v(plvp+i) = nodopa*i32 + nodob
i = i + 1
end if
nodopa = nodob
if (nodopa.ne.ksalva) go to 200
c
c puntlv: pointer at the first empty position of v()
c punta : pointer at the active problem with best lb
c puntb : pointer at the active problem with worst lb
c
ncodal = ncodal + 1
if (ncodal.eq.1) then
c
c insert the first problem
c
punta = puntlv
puntb = puntlv + 1
else
sc2 = v(punta+ioffv)
sc3 = v(puntb+ioffv-1)
lb = totass
if (lb.lt.v(punta+2) .or. (lb.eq.v(punta+2).and.(sc1.ge.sc2)))
+ then
c
v(puntlv) = punta
v(punta+1) = puntlv + 1
punta = puntlv
else if (lb.gt.v(puntb+1) .or. (lb.eq.v(puntb+1).and.
+ (sc1.le.sc3))) then
c
v(puntlv+1) = puntb
v(puntb-1) = puntlv
puntb = puntlv + 1
else
c
c find the first active problem with cost greater than lb
c
pcorr = v(punta)
220 sc2 = v(pcorr+ioffv)
lb1 = v(pcorr+2)
if ((lb.lt.lb1) .or. ((lb.eq.lb1).and. (sc1.ge.sc2))) then
c
c insert the problem in queue
c
puntol = v(pcorr+1) - 1
v(puntol) = puntlv
v(puntlv) = pcorr
v(pcorr+1) = puntlv + 1
v(puntlv+1) = puntol + 1
else
pcorr = v(pcorr)
go to 220
end if
end if
end if
puntlv = puntlv + ioffv + nnodi + 1
return
end
c *************************************************************************
subroutine exque(ordv,v,ordx,x,fmvf,vd,f,p2,ngen,punta,nprob,
+ iva,ivb,ncodal,n,sc1,totass,ioffv)
integer fmvf,ioffv,n,ncodal,ngen,nprob,ordv,ordx,p2,punta,sc1,
+ totass
integer f(n),iva(n),ivb(n),v(ordv),vd(n),x(ordx)
c .. local scalars r32,i,i32,ix,j,jp
c ..
i32 = 32000
r32 = 32000.
p2 = punta
ngen = float(v(punta+4))/r32
nprob = v(punta+4) - ngen*i32
totass = v(punta+2)
jp = v(punta+5)
j = 1
do 100 i = 1,nprob
ix = float(v(punta+ioffv+i))/r32
iva(j) = ix
ivb(j) = v(punta+ioffv+i) - ix*i32
j = j + 1
100 continue
c
c load the new vectors vd and f
c
do 200 i = 0,n - 1
vd(i+1) = x(jp+i)
f(i+1) = x(jp+n+i)
200 continue
c
x(jp) = fmvf
fmvf = jp
sc1 = v(punta+ioffv)
punta = v(punta)
ncodal = ncodal - 1
return
end
c *************************************************************************
subroutine actpro(patt,n,nprobv,pnuovo,ngen,vimpa,vimpb,ordv,v,cr,
+ ordsp,ic,c,ioffv,inf)
integer ordsp,ordv,patt,pnuovo,vimpa,vimpb
integer c(ordsp),cr(n+1),ic(ordsp),v(ordv)
integer p1,pold,pp
c subroutines ..modmat
m1 = -1
p1 = 1
i32 = 32000
r32 = 32000.
c update the cost matrix from the actual one(associated with patt)
c to the new one(associated with pnuovo)
if (v(pnuovo+3).ne.patt) then
c the problem patt is not the father of pnuovo
c
c "mark" the street from pnuovo to the root node
pp = pnuovo
50 if (pp.eq.0) then
ngenol = nprobv
c find a marked problem
60 nodoa = float(v(patt + ioffv + ngenol))/r32
nodob = v(patt + ioffv + ngenol) - nodoa*i32
nodoc = vimpa
nodod = vimpb
c update the matrix from the one associated with the son of patt
c to the one associated with patt
call modmat(cr,ordsp,ic,c,nodoa,nodob,m1,inf)
if (nodoc.ne.0) call modmat(cr,ordsp,ic,c,nodoc,nodod,m1,
+ inf)
if (v(patt+5).le.0) then
c the actual matrix (associated with the problem patt)
c is the one associated with common predecessor of pnew
c and the first patt
c
c pold : common predecessor problem
pp = pnuovo
pold = patt
c
70 patt = v(pp+3)
ngenpp = float(v(pp+4))/r32
c prohibit arcs
nodoa = float(v(patt + ioffv + ngenpp))/r32
nodob = v(patt + ioffv + ngenpp) - nodoa*i32
nodoc = float(v(pp+6))/r32
nodod = v(pp+6) - nodoc*i32
call modmat(cr,ordsp,ic,c,nodoa,nodob,p1,inf)
if (nodoc.ne.0) call modmat(cr,ordsp,ic,c,nodoc,nodod,
+ p1,inf)
if (patt.eq.pold) then
c
c update the marked problems
c
pp = pnuovo
75 v(pp+5) = -v(pp+5)
pp = v(pp+3)
if (pp.ne.0) go to 75
else
pp = patt
go to 70
end if
else
ngenol = float(v(patt+4))/r32
vimpa = float(v(patt+6))/r32
vimpb = v(patt+6) - vimpa*i32
patt = v(patt+3)
go to 60
end if
else
c
v(pp+5) = -v(pp+5)
pp = v(pp+3)
go to 50
end if
else
c
c the problem patt is the father of pnuovo
c
if (ngen.eq.nprobv) return
nodoa = float(v(patt + ioffv + nprobv))/r32
nodob = v(patt + ioffv + nprobv) - nodoa*i32
c
c update the cost of the prohibited arc
c
call modmat(cr,ordsp,ic,c,nodoa,nodob,m1,inf)
nodoc = vimpa
nodod = vimpb
c update the cost of the ''implicit'' prohibited arc
if (nodoc.ne.0) call modmat(cr,ordsp,ic,c,nodoc,nodod,m1,
+ inf)
nodoa = float(v(patt + ioffv + ngen))/r32
nodob = v(patt + ioffv + ngen) - nodoa*i32
c prohibit the arc nodoa,nodob
call modmat(cr,ordsp,ic,c,nodoa,nodob,p1,inf)
nodoc = float(v(pnuovo+6))/r32
nodod = v(pnuovo+6) - nodoc*i32
c prohibit the ''implicit''arc
if (nodoc.ne.0) call modmat(cr,ordsp,ic,c,nodoc,nodod,p1,
+ inf)
return
end if
return
end
c *************************************************************************
subroutine genson(n,iva,ivb,vd,nprob,i,vimpa,vimpb,nodoim,cr,
+ ordsp,ic,c,inf)
integer i,inf,n,nodoim,nprob,ordsp,vimpa,vimpb
integer c(ordsp),cr(n+1),ic(ordsp),iva(n),ivb(n),vd(n)
integer i2,ibm1,im1,m1,p1
c subroutines modmat
m1 = -1
p1 = 1
vimpa = 0
vimpb = 0
c prohibit i-th arc
call modmat(cr,ordsp,ic,c,iva(i),ivb(i),p1,inf)
if (i.ne.1) then
im1 = i - 1
ibm1 = ivb(im1)
i2 = ibm1
call modmat(cr,ordsp,ic,c,iva(im1),i2,m1,inf)
vd(ibm1) = vd(ibm1) - inf
c don't prohibit 2 times the same arc
if (nodoim.ne.ivb(i)) then
i2 = nodoim
call modmat(cr,ordsp,ic,c,iva(i),i2,p1,inf)
vimpa = iva(i)
vimpb = nodoim
else
vimpa = 0
vimpb = 0
end if
else
nodoim = ivb(nprob)
return
end if
c
c if necessary adjust the 'implicit' prohibited arc
c
i2 = nodoim
if (i.ge.3) call modmat(cr,ordsp,ic,c,iva(im1),i2,m1,inf)
return
end
c *************************************************************************
subroutine calcur(n,cr,ordsp,ic,c,f,ivai,vd,ur,inf)
c compute the value of the dual variable associated with row ivai
integer inf,ivai,n,ordsp,ur
integer c(ordsp),cr(n+1),f(n),ic(ordsp),vd(n)
integer jk,kl,pr,ul,vr
c ..
jk = f(ivai)
vr = vd(jk)
pr = cr(ivai)
ul = cr(ivai+1) - 1
do 100 kl = pr,ul
if (ic(kl).eq.jk) then
ur = c(kl) - vr - inf
go to 200
end if
100 continue
200 return
end
c *************************************************************************
subroutine calcud(n,ordsp,ic,cr,c,vc,fc,uc)
c compute the current dual variables u
integer n,ordsp
integer c(ordsp),cr(n+1),fc(n),ic(ordsp),uc(n),vc(n)
integer i,ifs,ii,ils,jk
c ..
do 100 i = 1,n
jk = fc(i)
ifs = cr(i)
ils = cr(i+1) - 1
do 50 ii = ifs,ils
if (ic(ii).eq.jk) uc(i) = c(ii) - vc(jk)
50 continue
100 continue
return
end
c *************************************************************************
subroutine modmat(cr,ordsp,ic,c,nodoa,nodob,flag,inf)
c if flag=1 exclude arc (nodoa,nodob)
c if flag=-1 remove exclusion constraint form arc (nodoa,nodob)
integer flag,inf,nodoa,nodob,ordsp
integer c(ordsp),cr(*),ic(ordsp)
integer j,k,pr,ul
c ..
pr = cr(nodoa)
ul = cr(nodoa+1) - 1
do 100 j = pr,ul
k = ic(j)
if (k.eq.nodob) then
c(j) = c(j) + flag*inf
go to 200
end if
100 continue
200 return
end
c *************************************************************************
subroutine creams(gap,a,n,cr,x,mm6,mm4,mm3,u,v,zeri,nlsp,err)
c reduction procedure
c store the sparse matrix in x() and defines mm3,mm4
c vector ic() is stored from mm3 and vector c() from mm4
integer err,gap,mm3,mm4,mm6,n,nlsp,zeri
integer a(n,n),cr(n+1),u(n),v(n),x(*)
integer gap2,i,ia,iu,j,k,middle
c ..
zeri = 0
err = 0
c
mm4 = mm6
middle = (mm4+mm3)/2.
k = middle
do 100 i = n,1,-1
cr(i+1) = mm4
iu = u(i)
gap2 = gap + iu
do 50 j = n,1,-1
if (i.ne.j) then
ia = a(i,j) - v(j)
a(i,j) = ia - iu
if (ia.le.gap2) then
if (ia.eq.iu) zeri = zeri + 1
mm4 = mm4 - 1
if (mm4.le.middle) go to 500
x(k) = j
x(mm4) = ia - u(i)
k = k - 1
end if
end if
50 continue
100 continue
cr(1) = mm4
do 200 i = 1,n + 1
cr(i) = cr(i) - mm4 + 1
200 continue
nlsp = cr(n+1) - cr(1)
k = middle
mm3 = mm4 - 1
do 300 i = nlsp,1,-1
x(mm3) = x(k)
k = k - 1
mm3 = mm3 - 1
300 continue
mm3 = mm3 + 1
do 400 i = 1,n
u(i) = 0
v(i) = 0
400 continue
return
500 err = 1
return
end
c *************************************************************************
subroutine errors(outerr,err)
integer err,outerr
outerr = 1
if (err.ne.-1) then
if (err.eq.1) call out('Insufficient memory, increase nstac')
if (err.eq.2) call out('Insufficient memory, increase nstac')
if (err.eq.3) call out('Insufficient memory, increase nstac')
if (err.eq.4) call out('Insufficient memory, increase nstac')
if (err.eq.5) call out('Insufficient memory, increase nstac')
if (err.eq.6) then
call out('solution not optimal. increase alpha')
return
end if
else
call out('maxnd nodes explored.solution not optimal')
return
end if
call out('increase ordx')
return
end
c *************************************************************************
subroutine setupo(n,a,u,v,m,ic,cr,kth,maxdim,iflag,inf)
c define the sparse matrix corresponding to the complete cost matrix a.
c the sparse matrix is stored through vectors ic,cr and matrix a.
integer a(n,n),cr(n+1),ic(*),u(n),v(n)
c real alpha,am,dd,aisum,ps,qq,ra,sum,th
c integer i,ia,inf2,ips,j,jstep,l,lstep,nr
data qq/2./
c ..
inf2 = float(inf)/2.
iflag = 0
lstep = 10
jstep = n/lstep
if (n.lt.lstep) jstep = 1
nr = (n+jstep-1)/jstep
aisum = 0.0
do 100 j = 1,n,jstep
aisum = aisum - v(j)
100 continue
aisum = aisum*n
do 200 i = 1,n
do 150 j = 1,n,jstep
ia = a(i,j)
if (ia.gt.inf2) then
aisum = aisum + u(i) + v(j)
else
aisum = aisum + ia
end if
150 continue
aisum = aisum - nr*u(i)
200 continue
sum = aisum
dd = n*nr - nr
alpha = sum/dd
am = m
ra = qq*alog(float(n))/am
kth = alpha*ra + 0.5
c
l = 1
do 300 i = 1,n
cr(i) = l
th = kth + u(i)
do 250 j = 1,n
if (a(i,j)-v(j).le.th) then
ic(l) = j
l = l + 1
end if
250 continue
if (l.gt.maxdim) go to 400
300 continue
cr(n+1) = l
ps = 0.005
if (n.le.950) ps = 0.01
if (n.le.450) ps = 0.02
if (n.le.250) ps = 0.03
ips = ps*float(n*n)
if (l.lt.ips) iflag = 1
return
400 cr(1) = -1
return
end
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