HDU 3435 A new Graph Game（最小费用流：有向环权值最小覆盖）

http://acm.hdu.edu.cn/showproblem.php?pid=3435

题意：
有n个点和m条边，你可以删去任意条边，使得所有点在一个哈密顿路径上，路径的权值得最小。

思路：

费用流，注意判断重边，否则会超时。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<queue>
  6 using namespace std;
  7 typedef long long LL;
  8
  9 const int maxn=2000+5;
 10 const int INF=0x3f3f3f3f;
 11
 12 int map[maxn][maxn];
 13
 14 struct Edge
 15 {
 16     int from, to, cap, flow, cost;
 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     int MincostMaxdflow(int s, int t, LL & cost)
 79     {
 80         int flow = 0; cost = 0;
 81         while (BellmanFord(s, t, flow, cost) );
 82         return flow;
 83     }
 84 }t;
 85
 86 int n,m;
 87
 88 int main()
 89 {
 90     //freopen("D:\\input.txt", "r", stdin);
 91     int T;
 92     int kase=0;
 93     scanf("%d",&T);
 94     int u,v,d;
 95     while(T--)
 96     {
 97         memset(map,0,sizeof(map));
 98         scanf("%d%d",&n,&m);
 99         int src=0,dst=2*n+1;
100         t.init(dst+1);
101         for(int i=1;i<=n;i++)
102         {
103             t.AddEdge(src,i,1,0);
104             t.AddEdge(i+n,dst,1,0);
105         }
106         for(int i=0;i<m;i++)
107         {
108             scanf("%d%d%d",&u,&v,&d);
109             if(map[u][v]==0 || map[u][v]>d)
110             {
111                 t.AddEdge(u,v+n,1,d);
112                 t.AddEdge(v,u+n,1,d);
113                 map[u][v]=map[v][u]=d;
114             }
115         }
116         long long cost;
117         int flow=t.MincostMaxdflow(src,dst,cost);
118         printf("Case %d: ",++kase);
119         if(flow==n)  printf("%d\n",cost);
120         else printf("NO\n");
121     }
122     return 0;
123 }