题目传送门
1 /*
2 题意：每次能消灭一行或一列的障碍物，要求最少的次数。
3 匈牙利算法：把行和列看做两个集合，当有障碍物连接时连一条边，问题转换为最小点覆盖数==二分图最大匹配数
4 趣味入门：http://blog.csdn.net/dark_scope/article/details/8880547
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cstring>
9 #include <vector>
10 using namespace std;
11
12 const int MAXN = 5e2 + 10;
13 const int INF = 0x3f3f3f3f;
14 vector<int> G[MAXN];
15 bool vis[MAXN];
16 int lk[MAXN];
17
18 bool DFS(int u)
19 {
20 for (int i=0; i<G[u].size (); ++i)
21 {
22 int v = G[u][i];
23 if (!vis[v])
24 {
25 vis[v] = true;
26 if (lk[v] == -1 || DFS (lk[v]))
27 {
28 lk[v] = u; return true;
29 }
30 }
31 }
32 return false;
33 }
34
35 int hungary(int n)
36 {
37 int res = 0; memset (lk, -1, sizeof (lk));
38 for (int i=1; i<=n; ++i)
39 {
40 memset (vis, false, sizeof (vis));
41 if (DFS (i)) res++;
42 }
43
44 return res;
45 }
46
47 int main(void) //POJ 3041 Asteroids
48 {
49 // freopen ("POJ_3041.in", "r", stdin);
50
51 int n, k;
52 while (scanf ("%d%d", &n, &k) == 2)
53 {
54 for (int i=1; i<=n; ++i) G[i].clear ();
55 for (int i=1; i<=k; ++i)
56 {
57 int x, y; scanf ("%d%d", &x, &y);
58 G[x].push_back (y);
59 }
60
61 printf ("%d
", hungary (n));
62 }
63
64 return 0;
65 }