POJ 3041Asteroids 【二分图】【最小点覆盖】

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此算法解决了一个问题：如何使用最少的射击次数来清除N×N网格中所有的小行星，每次射击可以清除一行或一列的所有小行星。通过构建二分图匹配算法找到最优解。

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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25922		Accepted: 14008

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

Sample Output

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).

#include<cstdio>
#include<cstring>
using namespace std;
const int MAX = 505;
int line[MAX][MAX], used[MAX], nex[MAX], n, m;
bool dfs(int x){
    for(int i = 1; i <= n; i++)
    if(line[x][i] && !used[i]){
        used[i] = 1;
        if(nex[i] == 0 || dfs(nex[i])){
            nex[i] = x;
            return 1;
        }
    }
    return 0;
}
int match(){
    int sum = 0;
    for(int i = 1; i <= n; i++){
        memset(used, 0, sizeof used);
        if(dfs(i)) sum++;
    }
    return sum;
}
int main(){
    while(scanf("%d%d", &n, &m) != EOF){
        memset(line, 0, sizeof line);
        memset(nex, 0, sizeof nex);
        for(int i = 0, x, y; i < m; i++){
            scanf("%d%d", &x, &y);
            line[x][y] = 1;
        }
        printf("%d\n", match());
    }
    return 0;
}