poj 3041 最小覆盖数等于最大匹配数

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12768		Accepted: 6949

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

 

建模:各行转为X点集合各点，各列转为Y点集合中各点，然后就是求最小覆盖，即选最少的点，使得每条边都与选出来的点关联。

关键是证明最小覆盖数等于最大匹配数，matrix67:http://www.matrix67.com/blog/archives/116 的证明据说很简单，但似没看懂。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

#define MAXN 555
#define MAXM 11111
int g[MAXN][MAXN], lft[MAXN];
bool t[MAXN];
int n;

bool match(int u){
    for(int v=1; v<=n; v++) if(g[u][v] && !t[v]){
        t[v]=true;
        if(lft[v]==-1 || match(lft[v])){
            lft[v]=u;
            return true;
        }
    }
    return false;
}

int main(){
//  freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int m;
    while(scanf(" %d %d", &n, &m)==2){
        int i, u, v;
        memset(g, 0, sizeof(g));
        for(i=0; i<m; i++){
            scanf(" %d %d", &u, &v);
            g[u][v]=1;
        }
        int ans=0;
        memset(lft, -1, sizeof(lft));
        for(i=1; i<=n; i++){
            memset(t, 0, sizeof(t));
            if(match(i)) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}