POJ3041(最小顶点覆盖)

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18611		Accepted: 10134

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
思路:把光束当作图的顶点，而把小行星当作光束的边。光束的方案为求一个顶点集合S，使得图中所有的边的一端至少有一个顶点在S中，即最小顶点覆盖问题。在二分图中，|最小顶点覆盖|=|最大匹配|

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=1005;
int n,m;
vector<int> arc[MAXN];
int match[MAXN],vis[MAXN];
bool dfs(int u)
{
    for(int i=0;i<arc[u].size();i++)
    {
        int to=arc[u][i];
        if(!vis[to])
        {
            vis[to]=1;
            int w=match[to];
            if(w==-1||dfs(w))
            {
                match[u]=to;
                match[to]=u;
                return true;
            }
        }
    }
    return false;
}
int max_flow()
{
    int ans=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<=n;i++)
    {
        if(match[i]==-1)
        {
            memset(vis,0,sizeof(vis));
            if(dfs(i))    ans++;
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)    arc[i].clear();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            v+=n;
            arc[u].push_back(v);
            arc[v].push_back(u);
        }
        n+=n;
        int res=max_flow();
        printf("%d
",res);
    }
    return 0;
}