POJ 1274 The Perfect Stall(二分图 && 匈牙利 && 最小点覆盖)
嗯…
题目链接:http://poj.org/problem?id=1274
一道很经典的匈牙利算法的题目:
将每只奶牛看成二分图中左边的点,将牛圈看成二分图中右边的点,如果奶牛看上某个牛圈,就将两点之间连边,然后跑一边匈牙利就行了…
AC代码:
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4
5 using namespace std;
6
7 int n, m, match[205], g[205][205], vis[205];
8
9 inline int dfs(int u){
10 for(int i = 1; i <= m; i++){
11 if(g[u][i] && !vis[i]){
12 vis[i] = 1;
13 if(!match[i] || dfs(match[i])){
14 match[i] = u;
15 return 1;
16 }
17 }
18 }
19 return 0;
20 }
21
22 inline int hungary(){
23 int ans = 0;
24 for(int i = 1; i <= n; i++){
25 memset(vis, 0, sizeof(vis));
26 if(dfs(i)) ans++;
27 }
28 return ans;
29 }
30
31 int main(){
32 while(~scanf("%d%d", &n, &m)){
33 memset(g, 0, sizeof(g));
34 memset(match, 0, sizeof(match));
35 for(int i = 1; i <= n; i++){
36 int s;
37 scanf("%d", &s);
38 for(int j = 1; j <= s; j++){
39 int t;
40 scanf("%d", &t);
41 g[i][t] = 1;
42 }
43 }
44 printf("%d\n", hungary());
45 }
46 return 0;
47 }
AC代码
转载于//www.cnblogs.com/New-ljx/p/11420923.html
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