二分图最小点覆盖数—

二分图最小点覆盖数——POJ 3041

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对应POJ题目：点击打开链接

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15911		Accepted: 8662

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

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is an asteroid and "." is empty space:
X.X .X. .X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (1,1) and (1,3), and then she may fire down column 2 to destroy the asteroids at (2,2) and (3,2).

Source

题意：在一个N*N的图里，散布着一些小行星，小明有一支很牛逼的枪，他一枪可以把某一行或某一列的小行星化为尘埃，问小明至少需要多少枪才能把所以的小行星消灭掉。

思路：把行放在X集合，列放在Y集合，那在某坐标的小行星就表示从X集合的某行连一条线到Y集合中的某列，一枪打在X集合里的某行，就表示该行所连的边也会被打到，Y集合的列同理。那问题就变为求二分图的最小点覆盖数了。。。

最小点覆盖数 = = 最大匹配数

假设最大匹配数为M，最小覆盖点数为N；

则每条边最少需要1个点关联，即N >= M；

又N <= M；因为只需要每个点分别关联每条匹配边，那其他边肯定会被某些点关联到，假设还存在某条边没有被关联到，就说明还没有达到最大匹配。

所以N == M；

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#define ms(x,y) memset(x,y,sizeof(x))
const int MAXN=500+10;
const int INF=1<<30;
using namespace std;
bool net[MAXN][MAXN];
bool vis[MAXN];
int link[MAXN];
int n,m;bool dfs(int u)
{for(int v=1; v<=n; v++){if(!vis[v] && net[u][v]){vis[v] = 1;if(link[v] == -1 || dfs(link[v])){link[v] = u;return 1;}}}return 0;
}int MaxMatch()
{int num = 0;ms(link,-1);for(int u=1; u<=n; u++){ms(vis,0);if(dfs(u)) num++;}return num;
}int main()
{//freopen("in.txt","r",stdin);scanf("%d%d", &n,&m);for(int i=1; i<=m; i++){int u,v;scanf("%d%d", &u,&v);net[u][v] = 1;}int ans = MaxMatch();printf("%d\n", ans);return 0;
}

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