poj3041 二分图最小顶点覆盖
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如题:http://poj.org/problem?id=3041
Description
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
* 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
Sample Input
3 41 11 32 23 2
Sample Output
2
Hint
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
思路:每一道光束摧毁一行或一列,我们将每一个给出的点作为一个顶点,相同横坐标或相同纵坐标就连一条线代表光束。最终要求的就是所有的边都是顶点关联。简化后就变成将横坐标看恒一个顶点集合,纵坐标一个顶点集合,从搜友给出的横纵坐标集合中,每一个横坐标连一条边到纵坐标,求最小顶点覆盖。
又因为二分图最小订点覆盖=最大匹配。问题解决
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 550
int N,K;
int G[MAXN][2*MAXN];
int vis[MAXN*2];
int match[MAXN*2];
int dfs(int v)
{
int i;
for(i=N+1;i<=2*N;i++)
{
if(!vis[i]&&G[v][i]==1)
{
vis[i]=v;
if(match[i]==-1||dfs(match[i]))
{
match[i]=v;
return 1;
}
}
}
return 0;
}
int main()
{
// freopen("C:\\1.txt","r",stdin);
cin>>N>>K;
int i;
for(i=0;i<K;i++)
{
int x,y;
scanf("%d%d",&x,&y);
G[x][y+N]=1;
}
memset(match,-1,sizeof(match));
int res=0;
for(i=1;i<=N;i++)
{
memset(vis,0,sizeof(vis));
if(dfs(i))
res++;
}
printf("%d\n",res);
return 0;
}
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