POJ 2135 Farm Tour [minimum cost maximum flow]

Test instructions

There are n points and m bars, allowing you to start from 1 to N and then back from N to 1, without requiring all points to pass, but each side can only walk once. The Edge is a non-forward edge.

Ask the shortest distance to walk.

At the beginning of the problem has not been engaged in the cost of the flow, and later made to come back to see, think of the plan is not difficult to build, because to ensure that each side can only walk once, then we split the side into two points, a starting point and the end, the capacity is 1, the weight is the length of the road. Then the two endpoints to the starting point of the connection capacity is 1 weight is 0 of the edge, the end point to two endpoints, respectively, the capacity is 1 weight is 0 of the edge, from the source point to 1 of the capacity 2 weight 0 of the edge, from N to the meeting point with a capacity of 2 weight 0 edge.

#include <stdio.h>#include<queue>#defineMAXN 55000#defineMAXM 20002*5#defineINF 0x3f3f3f3fusing namespacestd;//start number must be minimum, end number must be maximumBOOLVIS[MAXN];//whether the record in the SPFA is in the queuestructedge{Edge*next,*op;//op is pointing to the opposite side    intT,c,v;//T next point number, C capacity, V weight}ES[MAXM],*V[MAXN];//es edge static adjacency table, V-point numberintn,m,s,t,ec=-1;//s source point minimum, T meeting point Max, EC current edge numberintDEMOND[MAXN],SP[MAXN],PREV[MAXN];//record distance in SPSPFA, prev record previous point pathEdge *PATH[MAXN];//Synchronize records with Prev, record to previous edgevoidAddedge (intAintBintVintC=INF) {Edge E1={v[a],0, B,c,v},e2={v[b],0A0,-v}; es[++ec]=e1; v[a]=&Es[ec]; es[++ec]=e2; v[b]=&Es[ec]; V[a]->op=v[b]; v[b]->op=v[a];}voidinit () {intn,m; scanf ("%d%d",&n,&m); S=0; t=n+m*2+1; EC=-1; Addedge (S),1,0,2); Addedge (N,t,0,2); intA,b,c;  for(intI=1; i<=m;i++) {scanf ("%d%d%d",&a,&b,&c); Addedge (A,n+i,0,1); Addedge (B,n+i,0,1); Addedge (n+m+i,a,0,1); Addedge (n+m+i,b,0,1); Addedge (n+i,n+m+i,c,1); }}BOOLSPFA () {intu,v;  for(u=s;u<=t;u++) {Sp[u]=INF; } Queue<int>Q; Prev[s]=-1;    Q.push (S); Sp[s]=0; Vis[s]=1;  while(!Q.empty ()) {u=Q.front (); Vis[u]=0;        Q.pop ();  for(Edge *k=v[u];k;k=k->next) {v=k->T; if(k->c>0&&sp[u]+k->v<Sp[v]) {Sp[v]=sp[u]+k->v; PREV[V]=T; PATH[V]=K; if(vis[v]==0) {Vis[v]=1;                Q.push (v); }            }        }    }    returnsp[t]!=INF;}intargument () {inti,cost=inf,flow=0; Edge*e;  for(i=t;prev[i]!=-1; i=Prev[i]) {e=Path[i]; if(e->c<cost) cost=e->C; }     for(inti=t;prev[i]!=-1; i=Prev[i]) {e=Path[i]; E->c-=cost;e->op->c+=Cost ; Flow+=e->v*Cost ; }    returnflow;}intMaxcostflow () {intflow=0;  while(SPFA ()) {Flow+=argument (); }    returnFlow;}intMain () {init (); printf ("%d\n", Maxcostflow ()); return 0;}

POJ 2135 Farm Tour [minimum cost maximum flow]