poj 3041Asteroids

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid.

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space.
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

Sample Output

2


二分匹配模板题。。求武器和敌人之间的最大匹配。

View Code

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int N=505;
 7 int map[N][N],vis[N];
 8 int mx[N],my[N];
 9 int n,m;
10 int dfs(int u)
11 {
12     int i;
13     for(i=1;i<=n;i++)
14     {
15         if(!vis[i]&&map[u][i])
16         {
17             vis[i]=1;
18             if(my[i]==-1||dfs(my[i]))
19             {
20                 my[i]=u;
21                 mx[u]=i;
22                 return 1;
23             }
24         }
25     }
26     return 0;
27 }
28 int hungary()
29 {
30     int i,cnt=0;
31     for(i=1;i<=n;i++)
32         if(mx[i]==-1)
33         {
34             memset(vis,0,sizeof(vis));
35             if(dfs(i)) cnt++;
36         }
37     return cnt;
38 }
39 int main()
40 {
41     int i,j,u,v;
42     scanf("%d%d",&n,&m);
43     memset(map,0,sizeof(map));
44     memset(mx,-1,sizeof(mx));
45     memset(my,-1,sizeof(my));
46     for(i=1;i<=m;i++)
47     {
48         scanf("%d%d",&u,&v);
49         map[u][v]=1;
50     }
51     printf("%d\n",hungary());
52     return 0;
53 }