POJ 3041 Asteroids 匈牙利算法

POJ 3041：http://poj.org/problem?id=3041

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20152 Accepted: 10934

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 41 11 32 23 2

Sample Output

2

题目思路：将所有x行视为一个点集，所有y列视为一个点集，那么（x，y）就表示x和y之间有一条边了。而这题所求是最小点覆盖，即最大匹配。

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int n,map[510][510],flag[510],vis[510];int find(int k);int main(){int i,j,k,sum,m,x,y;memset(map,0,sizeof(map));memset(flag,0,sizeof(flag));scanf("%d%d",&n,&m);for(i=1;i<=m;i++){scanf("%d%d",&x,&y);map[x][y]=1;}sum=0;for(i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(find(i))sum+=1;}printf("%d\n",sum);return 0;}int find(int k){int i,j;for(j=1;j<=n;j++){if(map[k][j] && !vis[j]){vis[j]=1;if(!flag[j] || find(flag[j])){flag[j]=k;return 1;}}}return 0;}