poj_3041

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13502 Accepted: 7351

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 41 11 32 23 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).

左边是行，右边是列，1,1表示第一行到第一列有边，

二分图匹配，最小点覆盖数：最小覆盖要求用最少的点让每条边都至少和其中一个点关联。另外，最小点覆盖数 = 最大匹配

#include <iostream>#include <cstdio>#include <cstring>using namespace std;bool map[510][510];bool visit[510];int linker[510];int n;bool dfs(int num){    for (int i = 1; i <= n; i++)    {        if (!visit[i] && map[num][i])        {            visit[i] = 1;            if (linker[i] == -1 || dfs(linker[i]))            {                linker[i] = num;                return 1;            }        }    }    return 0;}int match(){    int res = 0;    memset(linker, -1, sizeof (linker));    for (int i = 1; i <= n; i++)    {        memset(visit, 0, sizeof (visit));        if (dfs(i))            res++;    }    return res;}int main(){    int t;    while (scanf("%d%d", &n, &t) != EOF)    {        memset(map, 0, sizeof(map));        while (t--)        {            int x, y;            scanf("%d%d", &x, &y);            map[x][y] = 1;        }        int res = match();        printf("%d\n", res);    }    return 0;}