POJ 2135 Farm Tour(网络流之费用流）

题目地址：POJ 2135

来回走一遍可以看成从源点到汇点走两遍。将每个点的流量设为1，就可以保证每条边不重复。然后跑一次费用流就行了。当流量到了2之后停止，输出此时的费用。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include<algorithm>

using namespace std;
const int INF=0x3f3f3f3f;
int head[1100], source, sink, cnt, cost, flow;
int d[1100], vis[1100], cur[1100];
struct node
{
    int u, v, cap, cost, next;
}edge[100000];
void add(int u, int v, int cap, int cost)
{
    edge[cnt].v=v;
    edge[cnt].cap=cap;
    edge[cnt].cost=cost;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cap=0;
    edge[cnt].cost=-cost;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int spfa()
{
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));
    queue<int>q;
    q.push(source);
    d[source]=0;
    cur[source]=-1;
    int minflow=INF, i;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(d[v]>d[u]+edge[i].cost&&edge[i].cap)
            {
                d[v]=d[u]+edge[i].cost;
                minflow=min(minflow,edge[i].cap);
                cur[v]=i;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(d[sink]==INF) return 0;
    flow+=minflow;
    cost+=minflow*d[sink];
    //printf("%d\n",minflow);
    if(flow==2)
        return 0;
    for(i=cur[sink];i!=-1;i=cur[edge[i^1].v])
    {
        edge[i].cap-=minflow;
        edge[i^1].cap+=minflow;
    }
    return 1;
}
void mcmf()
{
    while(spfa()) ;
    printf("%d\n",cost);
}
int main()
{
    int n, m, i, u, v, w;
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    cnt=0;
    source=1;
    sink=n;
    cost=0;
    flow=0;
    while(m--)
    {
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,1,w);
        add(v,u,1,w);
    }
    mcmf();
    return 0;
}