poj3041

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12162		Accepted: 6620

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).

#include<iostream>

using namespace std;



int n,k;  //n矩阵规格，k星体数量

int V1,V2;       //二分图顶点集

 /*矩阵的行列分别属于二分图的两个顶点集V1、V2，其中行x∈V1，列y∈V2*/

bool grid[501][501];  //存储数据方式：可达矩阵

bool vis[501];     //记录V2的点每个点是否已被搜索过

int link[501];     //记录 V2中的点y 在 V1中 所匹配的点x的编号

int m;  //最大匹配数



/*Hungary Algorithm*/



bool dfs(int x)

{

	for(int y=1;y<=V2;y++)

		if(grid[x][y] && !vis[y])  //x到y相邻(有边) 且 节点y未被搜索

		{

			vis[y]=true;   //标记节点y已被搜索

			if(link[y]==0 || dfs(link[y])) //link[y]==0 : 如果y不属于前一个匹配M

			{                               //find(link[y] : 如果被y匹配到的节点可以寻找到增广路

				link[y]=x; //那么可以更新匹配M'(用M替代M')

				return true;  //返回匹配成功的标志

			}

		}

	return false;  //继续查找V1下一个x的邻接节点

}



void search(void)

{

	for(int x=1;x<=V1;x++)

	{

		memset(vis,false,sizeof(vis)); //清空上次搜索时的标记

		if(dfs(x))  //从V1中的节点x开始寻找增广路

			m++;

	}

	return;

}



int main(void)

{

	cin>>n>>k;

	V1=V2=n;



	int x,y;         //坐标(临时变量)

	for(int i=1;i<=k;i++)

	{

		cin>>x>>y;

		grid[x][y]=true;   //相邻节点标记

	}



	/*增广轨搜索*/



	search();



	/*Output*/



	cout<<m<<endl;



	return 0;

}