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  • 图论--网络流--最小费用流最大流模板

    
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #define INF 1e9
    using namespace std;
    const int maxn= 1000+10;
     
    struct Edge
    {
        int from,to,cap,flow,cost;
        Edge(){}
        Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}
    };
     
    struct MCMF
    {
        int n,m,s,t;
        vector<Edge> edges;
        vector<int> G[maxn];
        bool inq[maxn];
        int d[maxn];
        int p[maxn];
        int a[maxn];
     
        void init(int n,int s,int t)
        {
            this->n=n, this->s=s ,this->t=t;
            edges.clear();
            for(int i=0;i<n;++i) G[i].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 BellmanFord(int &flow,int &cost)
        {
            for(int i=0;i<n;++i) d[i]=INF;
            memset(inq,0,sizeof(inq));
            queue<int> Q;
            d[s]=0,inq[s]=true,a[s]=INF,p[s]=0;
            Q.push(s);
     
            while(!Q.empty())
            {
                int u=Q.front(); Q.pop();
                inq[u]=false;
     
                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]) { inq[e.to]=true; Q.push(e.to); }
                    }
                }
            }
            if(d[t]==INF) return false;
            flow += a[t];
            cost += a[t]*d[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 solve()
        {
            int flow=0,cost=0;
            while(BellmanFord(flow,cost));
            return cost;
        }
    }MM;
     
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)==2)
        {
            int src=0,dst=n+1;
            MM.init(n+2,src,dst);
            MM.AddEdge(src,1,2,0);
            MM.AddEdge(n,dst,2,0);
            while(m--)
            {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                MM.AddEdge(u,v,1,w);
                //MM.AddEdge(v,u,1,w);
            }
            printf("%d
    ",2*MM.solve());
        }
    }
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  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798626.html
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