hungarian

 1 int linker[MAXN * 2];
 2 bool used[MAXN * 2];
 3 
 4 bool dfs(int u) {
 5     for (int i = h[u]; i; i = e[i].n) {
 6         int v = e[i].t;
 7         if (!used[v]) {
 8             used[v] = 1;
 9             if (linker[v] == -1 || dfs(linker[v])) {
10                 linker[v] = u;
11                 return 1;
12             }
13         }
14     }
15     return 0;
16 }
17 
18 int hungarian(int n) {
19     int res = 0;
20     for (int i = 0; i <= n * 2; i++) {
21         linker[i] = -1;
22     }
23     for (int u = 1; u <= n; u++) {
24         for (int i = 1; i <= n * 2; i++) {
25             used[i] = 0;
26         }
27         if (dfs(u)) res++;
28     }
29     return res;
30 }
原文地址:https://www.cnblogs.com/Accpted/p/11191870.html