hdu3549Flow Problem【最大流】

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6817    Accepted Submission(s): 3178


Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
 

 

Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
 

 

Output
For each test cases, you should output the maximum flow from source 1 to sink N.
 

 

Sample Input
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
 

 

Sample Output
Case 1: 1 Case 2: 2
 

 

Author
HyperHexagon
 

 

Source
HyperHexagon's Summer Gift (Original tasks)
题意：最大流
分析：最大流模板
代码：
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <vector>
  6 using namespace std;
  7 
  8 const int maxn = 20  << 1;
  9 const int INF = 1000000000;
 10 
 11 struct Edge
 12 {
 13     int from, to, cap, flow;
 14 };
 15 
 16 struct Dinic
 17 {
 18     int n, m, s, t;
 19     vector<Edge> edges;
 20     vector<int>G[maxn];
 21     bool vis[maxn];
 22     int d[maxn];
 23     int cur[maxn];
 24 
 25     void ClearAll(int n) {
 26         for(int i = 0; i <= n; i++) {
 27             G[i].clear();
 28         }
 29         edges.clear();
 30     }
 31 
 32     void AddEdge(int from, int to, int cap) {
 33         edges.push_back((Edge){from, to, cap, 0} );
 34         edges.push_back((Edge){to, from, 0, 0} );
 35         m = edges.size();
 36         G[from].push_back(m - 2);
 37         G[to].push_back(m - 1);
 38         //printf("%din end\n",m);
 39     }
 40 
 41     bool BFS()
 42     {
 43         memset(vis, 0, sizeof(vis) );
 44         queue<int> Q;
 45         Q.push(s);
 46         vis[s] = 1;
 47         d[s] = 0;
 48         while(!Q.empty() ){
 49             int x = Q.front(); Q.pop();
 50             for(int i = 0; i < G[x].size(); i++) {
 51                 Edge& e = edges[G[x][i]];
 52                 if(!vis[e.to] && e.cap > e.flow) {
 53                     vis[e.to] = 1;
 54                     d[e.to] = d[x] + 1;
 55                     Q.push(e.to);
 56                 }
 57             }
 58         }
 59         return vis[t];
 60     }
 61 
 62     int DFS(int x, int a) {
 63         if(x == t || a == 0) return a;
 64         int flow = 0, f;
 65         for(int& i = cur[x]; i < G[x].size(); i++) {
 66             Edge& e = edges[G[x][i]];
 67             if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
 68                 e.flow += f;
 69                 edges[G[x][i]^1].flow -= f;
 70                 flow += f;
 71                 a -= f;
 72                 if(a == 0) break;
 73             }
 74         }
 75         return flow;
 76     }
 77 
 78     int Maxflow(int s, int t) {
 79         this -> s = s; this -> t = t;
 80         int flow = 0;
 81         while(BFS()) {
 82             memset(cur, 0, sizeof(cur) );
 83             flow += DFS(s, INF);
 84         }
 85         return flow;
 86     }
 87 };
 88 
 89 Dinic g;
 90 
 91 int main()
 92 {
 93     int t;
 94     scanf("%d",&t);
 95     int n, m;
 96     int a, b, c;
 97     for(int kase = 1; kase <= t; kase++) {
 98         scanf("%d %d",&n, &m);
 99         g.ClearAll(maxn);
100         for(int i = 0;i < m; i++ ) {
101             scanf("%d %d %d",&a, &b, &c);
102             g.AddEdge(a, b, c);
103         }
104         printf("Case %d: %d\n",kase, g.Maxflow(1, n) );
105     }
106     return 0;
107 }
View Code

 

posted @ 2014-05-27 12:41  悠悠我心。  阅读(276)  评论(0)  编辑  收藏  举报