POJ 3041 Asteroids ———— 匈牙利算法模板題 （最小點覆蓋）

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).

 

思路：

這是一道最小點覆蓋問題， 匈牙利算法模板題。

（最小點覆蓋問題 = 最大匹配問題）

剛開始根本沒看出來和匈牙利算法有啥關係。

將行號看爲左半邊的節點，將列號看爲右半邊的節點，若（x，y）有小行星，那麼鏈接左邊的x和右邊的y

然後就出現了一個二分圖，然後求它的最大匹配就行。

AC代碼:

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>

#define MAXN 505
#define INF 0x3f3f3f3f

using namespace std;

int n, k;
int match[MAXN];
bool edge[MAXN][MAXN], vis[MAXN];

int dfs(int u) {
	for (int i = 1; i <= n; i++) {
		if (vis[i] == false && edge[u][i] == true) {
			vis[i] = true;
			if (match[i] == 0 || dfs(match[i])) {
				match[i] = u;
				return 1;
			}
		}
	}
	return 0;
}

int main() {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= k; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		edge[a][b] = true;
	}
	int sum = 0;
	for (int i = 1; i <= n; i++) {
		memset(vis, false, sizeof(vis));
		if (dfs(i) == 1) {
			sum++;
		}
	}
	printf("%d\n", sum);

	return 0;
}