【POJ】3041 - Asteroids（最小点覆盖，好题）

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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20939		Accepted: 11373

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

3 4
1 1
1 3
2 2
3 2

Sample Output

2

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

USACO 2005 November Gold

这道题确实很巧，把问题转化成了二分图。

x和y分别是两个集合，然后每一个点的坐标（x，y）变成x到y的边，然后求最小点覆盖。

然后再根据最小点覆盖就是最大匹配数，用匈牙利算法就行了。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
bool vis[500+5];
int f[500+5];
vector<int> mapp[500+5];
bool find(int x)
{
	for (int i = 0 ; i < mapp[x].size() ; i++)
	{
		int y = mapp[x][i];
		if (vis[y])
			continue;		//搜过的不用再搜 
		vis[y] = true;
		if (f[y] == -1)		//没有匹配
		{
			f[y] = x;
			return true;
		}
		else if (find(f[y]))
		{
			f[y] = x;
			return true;
		}
	}
	return false;
}
int main()
{
	int n,m;
	CLR(f,-1);
	scanf ("%d %d",&n,&m);
	while (m--)
	{
		int x,y;
		scanf ("%d %d",&x,&y);
		mapp[x].push_back(y);
	}
	int ant = 0;		//最大匹配 
	for (int i = 1 ; i <= n ; i++)
	{
		CLR(vis,false);
		if (find(i))
			ant++;
	}
	printf ("%d\n",ant);		//最大匹配即为最小点覆盖 
	return 0;
}