poj3041 二分图最小顶点覆盖

 

如题：http://poj.org/problem?id=3041

 

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16442		Accepted: 8967

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

 

 

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
#include
#include
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  {
  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;
}