最小费用最大流模版(训练指南)

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

struct MCMF
{
    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)
    {
        this->n=n;
        REP(i,1,n) G[i].clear();
        edges.clear();
    }
    void 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_ford(int s,int t,int &flow,int &cost)
    {
        REP(i,1,n) d[i]=INF;
        MS0(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]];
                inq[u]=0;
                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 0;
        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 1;
    }
    int Mincost(int s,int t)
    {
        int flow=0,cost=0;
        while(bellman_ford(s,t,flow,cost));
        return cost;
    }
};MCMF F;
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posted @ 2016-01-31 17:16  __560  阅读(269)  评论(0编辑  收藏  举报