POJ2135（最小费用最大流）

建立一个超级源点到1这个点的容量为2，费用为0，超级汇点同理，模版题。

#include <stdio.h>
#include <iostream>
#include <iostream>
#include <string>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;

const int maxn=1005*1005*2;
const int INF=999999999;

struct Edge
{
	int from,to,cap,flow,cost;
};

int n,m,s,t;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];

void init(int n)
{
	for(int i=0;i<=n+1;i++)
		G[i].clear();
	edges.clear();
}

void AddEdge(int from,int to,int cap,int cost)
{
	Edge e={from,to,cap,0,cost};
	edges.push_back(e);
	Edge p={to,from,0,0,-cost};
	edges.push_back(p);
	m=edges.size();
	G[from].push_back(m-2);
	G[to].push_back(m-1);
}

bool BellmanFord(int s,int t,int &flow,int &cost)
{
	for(int i=0;i<=n+1;i++)
	{
		d[i]=INF;
	}
	memset(inq,0,sizeof(inq));
	d[s]=0,inq[s]=1,p[s]=0,a[s]=INF;

	queue<int>q;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		inq[u]=0;
		for(int i=0;i<G[u].size();i++)
		{
			Edge &e=edges[G[u][i]];
			if(e.cap>e.flow && d[e.to]>d[u]+e.cost)
			{
				d[e.to]=d[u]+e.cost;
				p[e.to]=G[u][i];
				a[e.to]=min(a[u],e.cap-e.flow);
				if(!inq[e.to])
				{
					q.push(e.to);
					inq[e.to]=1;
				}
			}
		}
	}

	if(d[t]==INF)
		return false;
	flow+=a[t];
	cost+=(d[t]*a[t]);
	int u=t;
	while(u!=s)
	{
		edges[p[u]].flow+=a[t];
		edges[p[u]^1].flow-=a[t];
		u=edges[p[u]].from;
	}
	return true;
}

int Mincost(int s,int t)
{
	int flow=0,cost=0;
	while(BellmanFord(s,t,flow,cost));
	return cost;
}

int main()
{
	int k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		for(int i=0;i<k;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			AddEdge(u,v,1,w);
			AddEdge(v,u,1,w);
		}
		AddEdge(0,1,2,0);
		AddEdge(n,n+1,2,0);
		int ans=Mincost(0,n+1);
		printf("%d\n",ans);
	}
	return 0;
}

POJ2135（最小费用最大流）

原文：http://blog.csdn.net/mfmy_szw/article/details/19675085