2015年ACM训练二分图

                                                                                              A - Asteroids

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

该题的原理为二分图中输入点的个数和连线的情况，输出最大的匹配数

代码如下：

#include <iostream>
#include <cstring>
using namespace std;
int line[510][510],boy[510],used[510];
int n;
int find(int x)
{
    int i;
    for(i=1; i<=n; i++)
    {
        if(line[x][i]==1&&used[i]==0)
        {
            used[i]=1;
            if(boy[i]==0||find(boy[i]))
            {
                boy[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,k,x,y,sum;
    cin>>n>>k;
    memset(line,0,sizeof(line));
    memset(boy,0,sizeof(boy));
    memset(used,0,sizeof(used));
    for(i=0; i<k; i++)
    {
        cin>>x>>y;
        line[x][y]=1;
    }
    sum=0;
    for(i=1; i<=n; i++)
    {
        memset(used,0,sizeof(used));
        if(find(i))
        {
            sum++;
        }
    }
    cout<<sum<<endl;

return 0;
}