Asteroids(二分图匹配)

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).

题目大意:要求你横着或者竖着开枪消灭一行或者一列的飞船还是什么，求最小开枪次数

解题思路:这是二分图匹配的模板，主要是存下来二分图的模板，以后可以用，二分图关键是构建图形，然后把模板套上去就可以了

#include<iostream>  #include<cstdio>#include<stdio.h>#include<cstring>  #include<cstdio>  #include<climits>  #include<cmath> #include<vector>#include <bitset>#include<algorithm>  #include <queue>#include<map>using namespace std;vector<int> G[2500];int a[505], flag[505];int n, k;bool dfs(int x){int i;for (i = 0; i < G[x].size(); i++){if (a[G[x][i]]==0){a[G[x][i]] = 1;if (flag[G[x][i]]==0||dfs(flag[G[x][i]]))//是否这个点已经占据或者是否占据这个点的可以换一个找到未被占据的点{flag[G[x][i]] = x;//这个点被占据return true;}}}return false;}int main(){int i, sum, x, y;memset(flag, 0, sizeof(flag));//标记是否该点已经匹配cin >> n >> k;for (i = 1; i <= k; i++){cin >> x >> y;G[x].push_back(y);}sum = 0;for (i = 1; i <= n; i++){memset(a, 0, sizeof(a));if (dfs(i)){sum++;}}cout << sum << endl;}