Asteroids POJ 3041 匈牙利算法

最新推荐文章于 2020-08-21 20:27:19 发布

原创最新推荐文章于 2020-08-21 20:27:19 发布

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博客讲述Bessie驾驶太空船穿越N x N网格状危险小行星场，其武器可一次清除一行或一列小行星。需根据小行星位置，找出清除所有小行星的最少射击次数。还提到可将行和列分别看作二部图左右部，出现点的行与列连线求解。

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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

把行看成二部图的左部，列看成右部，出现的点行与列之间连线

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<queue>
#include<map>
#include<set>
#include<math.h>
#include<vector>
#include<set>
#define INF 0x3f3f3f3f
#define LL long long
#define mem(a, b) memset(a, b, sizeof(a))
#define N 500 
using namespace std;
int p, n, cas, vis[N], mach[N];
int head[N], Next[10001], ver[10001], tot;

void add(int x, int y) {
	ver[++tot]=y, Next[tot]=head[x], head[x]=tot;
}

bool dfs(int x) {
	for(int i=head[x], y; i; i=Next[i])
		if(!vis[y=ver[i]]) {
			vis[y]=1;
			if(!mach[y]||dfs(mach[y])){
				mach[y]=x; return 1;	
			}
		}
	return 0;
}

int main() {
	scanf("%d%d",&p, &n);
	for(int i=1; i<=n; i++) {
		int x, y;
		scanf("%d%d",&x, &y);
		add(x, y);
	}
	int ans=0;
	for(int i=1; i<=p; i++) {
		mem(vis, 0);
		if(dfs(i)) ans++;
	}
	cout<<ans;
}