poj-3041-Asteroids

Posted 2020-10-16 zhang--yd

poj-3041-Asteroids

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24115		Accepted: 13066

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

 

 

17919386

3041

Accepted

1408K

16MS

G++

721B

2017-12-04 21:25:29

 

 

使用匈牙利算法。

典型的匈牙利算法，建立二分图模型。

 

 

// poj 3041 hungary 

#include <cstdio>  
#include <cstring> 

const int MAXN = 500 + 10; 

int n, k, mp[MAXN][MAXN], vis[MAXN], match[MAXN]; 

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

int main(){
	freopen("in.txt", "r", stdin); 

	int num, x, y; 
	while(scanf("%d %d", &n, &k) != EOF){
		memset(mp, 0, sizeof(mp)); 

		for(int i=0; i<k; ++i){
			scanf("%d %d", &x, &y);
			mp[x][y] = 1;  
		} 

		num = 0; 
		memset(match, -1, sizeof(match)); 

		for(int i=1; i<=n; ++i){ 
			memset(vis, 0, sizeof(vis)); 
			if( dfs( i ) ){
				++num; 
			} 
		} 
		printf("%d\n", num );
	}
	return 0; 
}