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).

思路

　　這道題的建模很經典，我們將每一行看作是一個X結點，每一列看作是一個Y結點，每個目標對應一條邊，則擊毀所有目標且發射次數最少就意味著選取最少的點，使得任意一條邊至少有一個端點被選中。而這個對應著二分圖最小覆蓋概念的定義，由定理知二分圖最少點覆蓋數 = 最大匹配數，所以利用匈牙利算法解決該題。

　　AC代碼如下：

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

const int maxn = 510;
const int maxk = 10010;
int n,k;
int line[maxn][maxn];
bool used[maxn];
int next[maxn];

bool find (int x) {
    for (int i = 1; i <= n; i++) {
        if (line[x][i] &&!used[i]) {
            used[i] = true;
            if (!next[i] || find(next[i])) {
                next[i] = x;
                return true;
            }
        }
    }
    return false;
}

int match() {
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        memset(used, 0, sizeof(used));
        if (find(i)) sum++;
    }
    return sum;
}

int main(void) {
    int u,v;
    while(~scanf("%d%d", &n, &k)) {
        memset(line, 0, sizeof(line));
        memset(next, 0, sizeof(next));
        while(k--) {
            scanf("%d%d", &u, &v);
            line[u][v] = 1;
        }
        printf("%d\n", match());
    }
    return 0;
}

POJ #3041 Asteroids 3041 二分圖最小覆蓋 最大匹配 匈牙利算法