POJ 3041

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12894 Accepted: 7018

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

一个网格上有很多点，一些点上有目标，一次可以打掉一行或是一列，问你最少打几次能把目标打完。

本题可以用二分图模型解决，把每一行看为一个点，每一列看为一个点，对于(x,y)上有目标就连x，y，那么每条边表示一个目标，对二分图求最小点覆盖即可。

#include<cstdio>#include<cstring>#define maxn 509using namespace std;int a[maxn][maxn];int n,m;int cx[maxn],cy[maxn],vis[maxn];int path(int u){    int v;    for(v=1;v<=n;v++)    {        if(a[u][v]&&!vis[v])        {            vis[v]=1;            if(cy[v]==-1||path(cy[v]))            {                cy[v]=u;                cx[u]=v;                return 1;            }        }    }    return 0;}int match(){    int res=0;    memset(cx,-1,sizeof(cx));    memset(cy,-1,sizeof(cy));    for(int i=1;i<=n;i++)    {        if(cx[i]==-1)        {            memset(vis,0,sizeof(vis));            res+=path(i);        }    }    return res;}int main(){    scanf("%d%d",&n,&m);    for(int i=0;i<m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        a[x][y]=1;    }    printf("%d",match());}