【ssl1341】Asteroids【二分图】【匈牙利算法】【最大匹配问题】

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).

中文翻译:

给出一个矩阵，上面有敌人，每个子弹可以打出一横行或者一竖行，问最少用多少子弹消灭都有敌人，如：

X.X
.X.
.X.

x表示敌人，显然用两个子弹就可以解决所有敌人。
输入的是敌人的坐标。

分析

这题同样是匈牙利算法的模板，可以参考人员分配这题的分析。

真的只有输入不同哦。。

我用的是邻接表做法，也可以邻接矩阵。

上代码

喜闻乐见的代码

#include
#include
#include
#include
typedef long long ll;
using namespace std;

int n,m,link[201],tot,ans,hd[10001],cover[201];

struct node
{
	int to,next;
}a[10001];
 
void add(int x,int y)
{
	a[++tot]=(node){y,hd[x]};
	hd[x]=tot;
}

bool find(int x)
{
	for(int i=hd[x];i>0;i=a[i].next)
	{
		int j=a[i].to;
		if(cover[j]==0)
		{
			cover[j]=1;
			int q=link[j];
			link[j]=x;
			if(q==0||find(q))
			{
				return true;
			}
			link[j]=q;
		}
	}
	return false;
}
 
int main()
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
    	int x,y;
    	cin>>x>>y;
    	add(x,y);
    }
    for(int i=1;i<=n;i++)
    {
    	memset(cover,0,sizeof(cover));
    	find(i);
    }
    for(int i=1;i<=n;i++)
    {
    	if(link[i]!=0)
    	{
    		ans++;
    	}
    }
    cout<<ans;
	return 0;
}