• 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
    ",cost);
    120         else printf("NO
    ");
    121     }
    122     return 0;
    123 }
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  • 原文地址:https://www.cnblogs.com/zyb993963526/p/6697403.html
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