navigate nav mini resp not define 一個 present exp

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
題意：有一個n*n的矩陣,輸入k行數,每行數是一個點的坐標，一次性可以去掉矩陣一行或者一列的~~那個單詞沒查，反正就是去掉這些坐標點，問你最少的操作次數
思路：就是一個最小頂點覆蓋的模板題，和杭電的2119一樣。

#include<stdio.h>
#include<string.h>
#define N 550

int n,k;
int book[N],e[N][N],match[N];

int dfs(int u)
{
    int i;
    for(i = 1; i <= n; i ++)
    {
        
 
if(!book[i]&&e[u][i])
        {
            book[i] = 1;
            if(!match[i]||dfs(match[i]))
            {
                match[i] = u;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int t1,t2,sum,i,j;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        sum = 0;
        memset(e,0,sizeof(e));
        memset(match,0,sizeof(match));
        for(i = 1; i <= k; i ++)
        {
            scanf("%d%d",&t1,&t2);
            e[t1][t2] = 1;
        }
        for(i = 1; i <= n; i ++)
        {
            memset(book,0,sizeof(book));
            if(dfs(i))
                sum ++;
        }
        printf("%d\n",sum);
    }
    return 0;
}

【二分匹配入門專題1】G - Asteroids poj3041【最小頂點覆蓋】