POJ 2594 Treasure Exploration【传递闭包+最小路径覆盖】

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 using namespace std;
 6 
 7 const int maxn = 505;
 8 const int INF = 1000000000;
 9 
10 int n;
11 int vis[maxn];
12 int Link[maxn];
13 vector<int> G[maxn];
14 bool Find(int u) {
15     for(int i = 0; i < G[u].size(); i++) {
16         int v = G[u][i];
17         if(!vis[v]) {
18             vis[v] = 1;
19             if(Link[v] == -1 || Find(Link[v])) {
20                 Link[v] = u;
21                 return true;
22             }
23         }
24     }
25     return false;
26 }
27 
28 int solve() {
29     memset(Link, -1, sizeof(Link));
30     int cnt = 0;
31     for(int i = 1; i <= n; i++) {
32         if(G[i].size()) {
33             memset(vis, 0, sizeof(vis));
34             if(Find(i)) cnt++;
35         }
36     }
37     return cnt;
38 }
39 
40 int W[maxn][maxn];
41 int main() {
42     int m;
43     int a, b;
44     while(scanf("%d %d",&n, &m) && ( n + m) ) {
45         for(int i = 1; i <= n; i++) G[i].clear();
46         memset(W, 0, sizeof(W));
47         while(m--) {
48             scanf("%d %d",&a, &b);
49             W[a][b] = 1;
50         }
51         for(int k = 1; k <= n; k++) {
52             for(int i = 1; i <= n; i++) {
53                 if(W[i][k]) {
54                     for(int j = 1; j <= n; j++) {
55                         W[i][j] = W[i][j] || (W[i][k] && W[k][j]);
56                     }
57                 }
58             }
59         }
60         for(int i = 1; i <= n; i++) {
61             for(int j = 1; j <= n; j++) {
62                 if(W[i][j]) G[i].push_back(j);
63             }
64         }
65         printf("%d\n",n - solve());
66     }
67     return 0;
68 }