UVa 1658 海军上将（最小费用最大流）

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。

  1 #include <iostream>  
  2 #include <cstring>  
  3 #include <algorithm>   
  4 #include <vector>
  5 #include <queue>
  6 #include <cmath>
  7 using namespace std;
  8 
  9 const int maxn = 10000 + 5;
 10 const int INF = 0x3f3f3f3f;
 11 
 12 typedef long long LL;
 13 
 14 struct Edge{
 15     int from, to, cap, flow, cost;
 16 
 17     Edge(int u, int v, int c, int f, int w) :from(u), to(v), cap(c), flow(f), cost(w) {}
 18 };
 19 
 20 struct MCMF
 21 {
 22     int n, m;
 23     vector<Edge> edges;
 24     vector<int> G[maxn];
 25     int inq[maxn];
 26     int d[maxn];
 27     int p[maxn];
 28     int a[maxn];
 29 
 30     void init(int n)
 31     {
 32         this->n = n;
 33         for (int i = 0; i<n; i++) G[i].clear();
 34         edges.clear();
 35     }
 36 
 37     void AddEdge(int from, int to, int cap, int cost)
 38     {
 39         edges.push_back(Edge(from, to, cap, 0, cost));
 40         edges.push_back(Edge(to, from, 0, 0, -cost));
 41         m = edges.size();
 42         G[from].push_back(m - 2);
 43         G[to].push_back(m - 1);
 44     }
 45 
 46     bool BellmanFord(int s, int t, int &flow, LL & cost)
 47     {
 48         for (int i = 0; i<n; i++) d[i] = INF;
 49         memset(inq, 0, sizeof(inq));
 50         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
 51 
 52         queue<int> Q;
 53         Q.push(s);
 54         while (!Q.empty()){
 55             int u = Q.front(); Q.pop();
 56             inq[u] = 0;
 57             for (int i = 0; i<G[u].size(); i++){
 58                 Edge& e = edges[G[u][i]];
 59                 if (e.cap>e.flow && d[e.to]>d[u] + e.cost){
 60                     d[e.to] = d[u] + e.cost;
 61                     p[e.to] = G[u][i];
 62                     a[e.to] = min(a[u], e.cap - e.flow);
 63                     if (!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
 64                 }
 65             }
 66         }
 67         if (d[t] == INF) return false;
 68         flow += a[t];
 69         cost += (LL)d[t] * (LL)a[t];
 70         for (int u = t; u != s; u = edges[p[u]].from){
 71             edges[p[u]].flow += a[t];
 72             edges[p[u] ^ 1].flow -= a[t];
 73 
 74         }
 75         return true;
 76     }
 77 
 78     void MincostMaxdflow(int s, int t, int limit, LL & cost){
 79         int flow = 0; cost = 0;
 80         while (BellmanFord(s, t, flow, cost) && flow < limit);
 81         //return flow;
 82     }
 83 }t;
 84 
 85 int main()
 86 {
 87     //freopen("D:\input.txt", "r", stdin);
 88     int n, m;
 89     while (~scanf("%d%d", &n, &m))
 90     {
 91         t.init(2 * n - 2);
 92         for (int i = 2; i <= n - 1; i++)
 93         {
 94             t.AddEdge(i - 1, n - 2 + i, 1, 0);
 95         }
 96         for (int i = 0; i<m; i++)
 97         {
 98             int u, v, w;
 99             scanf("%d%d%d", &u, &v, &w);
100             v--;
101             if (u != 1 && u != n) u += n - 2;
102             else u--;
103             t.AddEdge(u, v, 1, w);
104         }
105         LL cost;
106         t.MincostMaxdflow(0, n - 1, 2, cost);
107         printf("%lld
", cost);
108     }
109     return 0;
110 }