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 =  + ;
 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 = ; 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, , cost));
         edges.push_back(Edge(to, from, , , -cost));
         m = edges.size();
         G[from].push_back(m - );
         G[to].push_back(m - );
     }

     bool BellmanFord(int s, int t, int &flow, LL & cost)
     {
         for (int i = ; i<n; i++) d[i] = INF;
         memset(inq, , sizeof(inq));
         d[s] = ; inq[s] = ; p[s] = ; a[s] = INF;

         queue<int> Q;
         Q.push(s);
         while (!Q.empty()){
             int u = Q.front(); Q.pop();
             inq[u] = ;
             for (int i = ; 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] = ; }
                 }
             }
         }
         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] ^ ].flow -= a[t];

         }
         return true;
     }

     void MincostMaxdflow(int s, int t, int limit, LL & cost){
         int flow = ; cost = ;
         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( * n - );
         for (int i = ; i <= n - ; i++)
         {
             t.AddEdge(i - , n -  + i, , );
         }
         for (int i = ; i<m; i++)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             v--;
             if (u !=  && u != n) u += n - ;
             else u--;
             t.AddEdge(u, v, , w);
         }
         LL cost;
         t.MincostMaxdflow(, n - , , cost);
         printf("%lld\n", cost);
     }
     return ;
 }