POJ 3041 Asteroids

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10888 Accepted: 5898

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

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

解题思路：首先将问题转化为求最小点集覆盖问题，就是说，将x,y坐标分别看做两个集合，把每个坐标看成一条对应的匹配边，在两个集合中各取数个点，使得所有边均被覆盖，求最少需要多少个点。其实等价于求最大匹配数问题。证明请看

http://www.matrix67.com/blog/archives/116

#include<iostream>using namespace std;int map[550][550],pre[550],m,n;bool used[550];bool find(int x){int i;for(i=1;i<=n;i++){if(!used[i]&&map[x][i]){used[i]=true;if(pre[i]==-1||find(pre[i])){pre[i]=x;return true;}}}return false;}int MMG(){int i,count=0;memset(pre,-1,sizeof(pre));for(i=1;i<=n;i++){memset(used,false,sizeof(used));if(find(i))count++;}return count;}int main(){int a,b,i;while(cin>>n>>m){memset(map,0,sizeof(map));for(i=0;i<m;i++){cin>>a>>b;map[a][b]=1;}cout<<MMG()<<endl;}return 0;}