蒟蒻zyk又来发水题题解了。。。

题目链接：http://poj.org/problem?id=2135

题意：无向边的最小费用最大流，注意要另建超级源点和超级汇点，加一条无向边相当于加4条有向边。

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

#define MAXV 10100
#define MAXE 1000100
#define INF 0x3f3f3f3f

using namespace std;

struct edge
{
    int u,v,w,cap,next;
}edges[MAXE];

int n,m;
int head[MAXV],nCount=-1;
int last[MAXV],dist[MAXV];
bool inQueue[MAXV];

void AddEdge(int U,int V,int W,int C)
{
    //正向边
    edges[++nCount].u=U;
    edges[nCount].v=V;
    edges[nCount].w=W;
    edges[nCount].cap=C;
    edges[nCount].next=head[U];
    head[U]=nCount;
    //反向边
    edges[++nCount].u=V;
    edges[nCount].v=U;
    edges[nCount].w=-W;
    edges[nCount].cap=0;
    edges[nCount].next=head[V];
    head[V]=nCount;
}

bool SPFA(int s,int t)
{
    queue<int>q;
    memset(last,-1,sizeof(last));
    memset(dist,INF,sizeof(dist));
    memset(inQueue,false,sizeof(inQueue));
    inQueue[s]=true;
    dist[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        inQueue[u]=false;
        for(int p=head[u];p!=-1;p=edges[p].next)
        {
            if(edges[p].cap>0)
            {
                int v=edges[p].v;
                if(dist[u]+edges[p].w<dist[v])
                {
                    dist[v]=dist[u]+edges[p].w;
                    last[v]=p;
                    if(!inQueue[v])
                    {
                        inQueue[v]=true;
                        q.push(v);
                    }
                }
            }
        }
    }
    return dist[t]<INF;
}

int MCMF(int s,int t) //求s到t的最小费用最大流
{
    int sumcost=0,sumflow=0,flow;
    while(SPFA(s,t))
    {
        flow=INF;
        for(int i=last[t];i!=-1;i=last[edges[i].u])
        {
            if(edges[i].cap<flow)
                flow=edges[i].cap;
        }
        for(int i=last[t];i!=-1;i=last[edges[i].u])
        {
            edges[i].cap-=flow; //正向边减去容量
            edges[i^1].cap+=flow; //反向边加容量
        }
        sumcost+=dist[t];
        sumflow+=flow;
    }
    return sumcost;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(head,-1,sizeof(head));
        AddEdge(0,1,0,2); //建立超级源点
        AddEdge(n,n+1,0,2); //建立超级汇点
        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            AddEdge(u,v,w,1);
            AddEdge(v,u,w,1);
        }
        printf("%d\n",MCMF(0,n+1));
    }
    return 0;
}

[POJ 2135]Farm Tour(最小费用最大流)