二分图最大匹配(匈牙利算法) POJ 3041 Asteroids

 

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.youkuaiyun.com/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\n", hungary (n));
62     }
63 
64     return 0;
65 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4652268.html