Poj 3041 Asteroids

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25270		Accepted: 13635

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

 

//题意：有一个n*n的矩阵，上面有m个黑格子
//        现在有一把刷子，可以将一行或一列上的黑格子刷成白色
//        问最少需要几刷子可以让矩阵变成全白
//        格子不是黑的就是白的 

//用横坐标和纵坐标建立二分图，如果有黑格子在(x,y)
//那么就让左侧x连向右侧y
//那么我们就可以得到同行和同列的格子的个数
//然后求最小点覆盖就可以了 

//最小点覆盖=最大匹配数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

inline int read()
{
    char c=getchar();int num=0;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*10+c-'0';
    return num;
}

const int M=1e4+5;

int n,m;
int match[M];
int vis[M],tim;
int head[M],num_edge;
struct Edge
{
    int v,nxt;
}edge[M];

void add_edge(int u,int v)
{
    edge[++num_edge].v=v;
    edge[num_edge].nxt=head[u];
    head[u]=num_edge;
}

bool dfs(int u)
{
    for(int i=head[u],v;i;i=edge[i].nxt)
    {
        v=edge[i].v;
        if(vis[v]==tim)
            continue;
        vis[v]=tim;
        if(!match[v]||dfs(match[v]))
        {
            match[v]=u;
            return 1;
        }
    }
    return 0;
}

int main()
{
    n=read(),m=read();
    for(int i=1,x,y;i<=m;++i)
    {
        x=read(),y=read();
        add_edge(x,y);
    }
    int ans=0;
    for(int i=1;i<=n;++i)
    {
        ++tim;
        ans+=dfs(i);
    }
    printf("%d",ans);
    return 0;
}