poj 3041 二分图

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16344		Accepted: 8906

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

每个点i（Ri,Ci）要么被x=Ri，要么被y=Ci直线所覆盖。将直线看作图的点，这样可以分为水平的和竖直的两类点，每个小行星对应图中一条边（一个水平坐标和一个竖直坐标确定了一个位置）。每条边只需要被一个顶点覆盖就好了，问题就转化为最小顶点覆盖，因为是二分图，所以求解最大匹配就可以了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

#define maxn 505
#define maxk 10005

int n, k;
vector<int> g[2*maxn];

int match[2*maxn];
int vis[2*maxn];

int dfs(int u)
{
    int v;
    for(int i = 0; i < g[u].size(); i++){
        v = g[u][i];

        if(vis[v]) continue;

        if(!match[v]){
            match[v] = u;
            match[u] = v;
            return 1;
        }
        else{
            vis[v] = 1;
            if(dfs(match[v])){
                match[v] = u;
                match[u] = v;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    while(~scanf("%d%d", &n, &k)){
        int r,c;
        for(int i = 0; i < 2*n+1; i++) g[i].clear();

        for(int i = 0; i < k; i++){
            scanf("%d%d", &r, &c);
            g[r].push_back(n+c);
            g[n+c].push_back(r);
        }

        memset(match, 0, sizeof(match));

        int ans = 0;
        for(int i = 1; i <= n; i++)
        if(!match[i]){
             memset(vis, 0, sizeof(vis));
             ans += dfs(i);
        }

        printf("%d\n", ans);
    }
    return 0;
}