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poj 3041(最大匹配问题)

2024-07-12 14:32:38   227人阅读

http://poj.org/problem?id=3041
Asteroids
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14476 Accepted: 7880

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. 

题目大意是导弹可以消灭直线上的障碍物，给出障碍物的位置，问最少需要多少导弹，把矩阵中的行放一边，列放另一边，按二分图做，就不是很难了，这一类的题目主要就是主函数那里有点不一样，其他套着模板来。

模板题

AC代码：

#include <stdio.h>
#include <string.h>
#define MAXN 510

int N,K,ans;
bool map[MAXN][MAXN];
int match[MAXN];
bool vis[MAXN];

bool find(int u)
{
	int i;
	for ( i=1; i<=N; i++)
	{
		if (map[u][i] && !vis[i] )
		{
			vis[i] = true;
			if ( match[i] == -1 || find(match[i]) )
			{
				match[i] = u;
				return true;
			}
		}
	}
	return false;
}

int main()
{
	int i,x,y;
	ans = 0;
	scanf("%d%d",&N,&K);
	memset(map,0,sizeof(map));
	memset(match,-1,sizeof(match));
	for (i=0; i<K; i++)
	{
		scanf("%d%d",&x,&y);
		map[x][y] = 1;
	}
	for (i=1; i<=N; i++)
	{
		memset(vis,0,sizeof(vis));
		if (find(i))
			ans++;
	}
	printf("%d\n",ans);
	return 0;
}


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