POJ2135 Farm Tour（最小费用最大流）

题目问的是从1到n再回到1边不重复走的最短路，本质是找1到n的两条路径不重复的尽量短的路。

 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<algorithm>
 using namespace std;
 #define INF (1<<30)
 #define MAXN 111
 #define MAXM 2222
 struct Edge{
     int u,v,cap,cost,next;
 }edge[MAXM];
 int head[MAXN];
 int NV,NE,vs,vt;

 void addEdge(int u,int v,int cap,int cost){
     edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
     edge[NE].next=head[u]; head[u]=NE++;
     edge[NE].u=v; edge[NE].v=u; edge[NE].cap=; edge[NE].cost=-cost;
     edge[NE].next=head[v]; head[v]=NE++;
 }
 bool vis[MAXN];
 int d[MAXN],pre[MAXN];
 bool SPFA(){
     for(int i=;i<NV;++i){
         vis[i]=;
         d[i]=INF;
     }
     vis[vs]=;
     d[vs]=;
     queue<int> que;
     que.push(vs);
     while(!que.empty()){
         int u=que.front(); que.pop();
         for(int i=head[u]; i!=-; i=edge[i].next){
             int v=edge[i].v;
             if(edge[i].cap && d[v]>d[u]+edge[i].cost){
                 d[v]=d[u]+edge[i].cost;
                 pre[v]=i;
                 if(!vis[v]){
                     vis[v]=;
                     que.push(v);
                 }
             }
         }
         vis[u]=;
     }
     return d[vt]!=INF;
 }
 int MCMF(){
     int res=;
     while(SPFA()){
         int flow=INF,cost=;
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             flow=min(flow,edge[pre[u]].cap);
         }
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             edge[pre[u]].cap-=flow;
             edge[pre[u]^].cap+=flow;
             cost+=flow*edge[pre[u]].cost;
         }
         res+=cost;
     }
     return res;
 }
 int main(){
     int n,m,a,b,c;
     scanf("%d%d",&n,&m);
     vs=; vt=n+; NV=vt+; NE=;
     memset(head,-,sizeof(head));
     while(m--){
         scanf("%d%d%d",&a,&b,&c);
         addEdge(a,b,,c);
         addEdge(b,a,,c);
     }
     addEdge(vs,,,);
     addEdge(n,vt,,);
     printf("%d",MCMF());
     return ;
 }