poj3041 二分图最小顶点覆盖

Asteroids

Accepted: td> 9375

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17237

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

 

 

（以行列分别作为两个顶点集V1、V2，其中| V1|=| V2|）

然后把每行x或者每列y看成一个点，而障碍物(x,y)可以看做连接x和y的边。按照这种思路构图后。问题就转化成为选择最少的一些点(x或y)，使得从这些点与所有的边相邻，其实这就是最小点覆盖问题。

 

再利用二分图最大匹配的König定理：

最小点覆盖数 = 最大匹配数

 

（PS：最小点覆盖：假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖图的所有的边。）

 

因此本题自然转化为求 二分图的最大匹配 问题

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int g[505][505:span style="color: #000000;">];
int link[505];
bool vis[505];
int tx,ty;
bool find(int u){
    for(int i=1;i<=ty;i++){
        if(!vis[i]&&g[u][i]){
            vis[i]=true;
            if(link[i]==-1||find(link[i])){
                link[i]=u;
                return true;
            }
        }

    }
    return false;
}
int solve(){
   int sum=0;
   memset(link,-1,sizeof(link));
   for(int i=1;i<=tx;i++){
    memset(vis,false,sizeof(vis));
    if(find(i))
        sum++;
   }
   return sum;
}
int main(){
   int n;
   while(scanf("%d%d",&tx,&n)!=EOF){
       memset(g,0,sizeof(g));
       ty=tx;
       int x,y;
       for(int i=1;i<=n;i++){
        scanf("%d%d",&x,&y);
        g[x][y]=1;
       }
    printf("%d\n",solve());
   }
   return 0;
}