最小费用最大流（Bellman-Ford找增广路）

示例题目： POJ2135

#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int MAXN=1e3+50;//点数的最大值
const int MAXM=4e5+50;//边数的最大值
const int INF=0x3f3f3f3f;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
};

struct MCMF
{
    int n,m;
    vector edges;
    vector<int> G[MAXN];
    int inq[MAXN];//是否在队列中
    int d[MAXN];//Bellman-Ford
    int p[MAXN];//上一条弧
    int a[MAXN];//可改进量

    void init(int n)
    {
        this->n=n;
        for(int i=0;ivoid addEdge(int from,int to,int cap,int cost)
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool Bellman_Frod(int s,int t,int& flow,LL& cost)
    {
        for(int i=0;imemset(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;iif(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+=(LL)d[t]*(LL)a[t];
        //cout<<"d[t]: "<
        for(int u=t;u!=s;u=edges[p[u]].from)
        {
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
        }
        return true;
    }

    //需要保证初始网络中没有负权圈
    int MincostMaxflow(int s,int t,LL& cost)
    {
        int flow=0;
        cost=0;
        while(Bellman_Frod(s,t,flow,cost));
        return flow;
    }
};

MCMF cf;

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        cf.init(n+2);
        int u,v,w;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            cf.addEdge(u,v,1,w);
            cf.addEdge(v,u,1,w);
        }
        cf.addEdge(0,1,2,0);
        cf.addEdge(n,n+1,2,0);
        LL cost;
        cf.MincostMaxflow(0,n+1,cost);
        printf("%lld\n",cost);
    }
    return 0;
}