UVa 1658 海军上将（最小费用最大流）

https://vjudge.net/problem/UVA-1658

题意：

给出一个v个点e条边的有向加权图，求1~v的两条不相交（除了起点和终点外公共点）的路径，使得权和最小。

思路：
把2到v-1的每个点拆分为两个节点，容量为1，也就是只可以用一次，费用为0，然后求1到v的流量为2的最小费用流。

#include <iostream>  
#include <cstring>  
#include <algorithm>   
#include <vector>
#include <queue>
#include <cmath>
using namespace std;

const int maxn = 10000 + 5;
const int INF = 0x3f3f3f3f;

typedef long long LL;

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

    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<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;
        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, LL & cost)
    {
        for (int i = 0; i<n; 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 += (LL)d[t] * (LL)a[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;
    }

    void MincostMaxdflow(int s, int t, int limit, LL & cost){
        int flow = 0; cost = 0;
        while (BellmanFord(s, t, flow, cost) && flow < limit);
        //return flow;
    }
}t;

int main()
{
    //freopen("D:\input.txt", "r", stdin);
    int n, m;
    while (~scanf("%d%d", &n, &m))
    {
        t.init(2 * n - 2);
        for (int i = 2; i <= n - 1; i++)
        {
            t.AddEdge(i - 1, n - 2 + i, 1, 0);
        }
        for (int i = 0; i<m; i++)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            v--;
            if (u != 1 && u != n) u += n - 2;
            else u--;
            t.AddEdge(u, v, 1, w);
        }
        LL cost;
        t.MincostMaxdflow(0, n - 1, 2, cost);
        printf("%lld
", cost);
    }
    return 0;
}