HDU 1853 MCMF

题意：给定一个有向带权图，使得每一个点都在一个环上，而且权之和最小。

分析：每个点在一个环上，入度 = 出度 = 1，拆点入点，出点，s到所有入点全部满载的最小费用MCMF;

#include <bits/stdc++.h>

using namespace std;

const int maxn = 105*2;
const int INF = 0x3f3f3f3f;

typedef pair<int,int> pii;

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

struct MCMF
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool 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; i < n; i++) 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 BellmanFord(int s, int t, int &flow, long long& cost)
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;i++)
            d[i] = INF;
        d[s] = 0;
        inq[s] = true;
        p[s] = 0;
        a[s] = INF;

        queue<int> Q;
        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])
                    {
                        Q.push(e.to);
                        inq[e.to] = true;
                    }
                }
            }
        }
        if(d[t] == INF) return false; //s-t 不连通，失败退出
        flow += a[t];
        cost += (long long)d[t] * (long long)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;
    }

    pair<long long,int>Mincost(int s, int t)
    {
        long long cost = 0;
        int flow = 0;
        while(BellmanFord(s, t, flow, cost));
        return pair<int,long long>{flow,cost};
    }
}sol;


int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF) {
        int s = 0,t=2*n+1;
        sol.init(2*n+2);

        for(int i=1;i<=n;i++)
            sol.AddEdge(s,i,1,0);

        for(int i=n+1;i<=2*n;i++)
            sol.AddEdge(i,t,1,0);

        for(int i=0;i<m;i++) {
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            sol.AddEdge(u,v+n,1,c);
        }

        pii ans = sol.Mincost(s,t);
        if(ans.first!=n)
            puts("-1");
        else cout<<ans.second<<endl;

    }
    return 0;
}