POJ3041 Asteroids【最小顶点覆盖问题】

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).

解题思路

题目大意：给一个N*N的矩阵，有些格子有障碍，要求我们消除这些障碍，问每次消除一行或一列的障碍，最少要几次。

思路：关键是构图，把每行看作X集合中的一个点，每列看作Y集合中的一个点，这样每个障碍的行号列号对应连接XY集合的一条边。现在要用最少点的点覆盖所有的障碍（也就是边），这就变成了最小顶点覆盖问题了。其中：最小顶点数 = 最大匹配。

代码：

#include
#include
#include
using namespace std;
int map[550][550];
int match[10010], vis[10010];
int n, k;

int dfs(int start)
{
	for (int i = 1; i <= n; i++)
	{
		if (vis[i] == 0 && map[start][i] == 1)
		{
			vis[i] = 1;
			if (match[i] == 0 || dfs(match[i]) == 1)
			{
				match[i] = start;
				return 1;
			}
		}
	}
	return 0;
}

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