【POJ 3041 Asteroids】+ 二分图

Asteroids
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7887 Accepted: 4196

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).

裸二分图，把x和y尽可能多的相匹配即是最优解~~~（POJ 不能用 < bits/stdc++.h >头文件，莫名的WA一次=-=）

AC代码 ：

#include<cstdio>#include<cstring>#include<vector>using namespace std;vector <int> v[510];int vis[1011],falg[1011],N,M,a,b;int DFS(int x){    for(int i = 0 ; i < v[x].size(); i++){        int u = v[x][i];        if(!vis[u]){            vis[u] = 1;            if(!falg[u] || DFS(falg[u])){                falg[u] = x;                return 1;            }        }    }    return 0;}int main(){    while(scanf("%d %d",&N,&M)!=EOF){        for(int i = 1 ; i <= N ; i++)            v[i].clear();        memset(falg,0,sizeof(falg));        while(M--){            scanf("%d %d",&a,&b);            v[a].push_back(b);        }        int cut = 0;        for(int i = 1 ; i <= N ; i++){            memset(vis,0,sizeof(vis));            cut += DFS(i);        }        printf("%d\n",cut);    }    return 0;}