【Asteroids】二分图

最新推荐文章于 2020-03-19 20:04:48 发布

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Bessie需穿越危险的陨石区，该区域为N×N网格，包含K颗陨石。她拥有能清除任意一行或一列陨石的强大武器，目标是最少使用次数清除所有陨石。

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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).

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k;
int coordinate[505][505];
int f[505];
int vis[505];
bool find(int x)
{
    for(int i = 1; i <= n; i++)
    {
        if(coordinate[x][i]&&!vis[i])
        {
            vis[i]=true;
            if(f[i]==-1)
            {
                f[i]=x;
                return true;
            }
            else if(find(f[i]))
            {
                f[i]=x;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    scanf("%d%d",&n,&k);
    memset(coordinate, false, sizeof(coordinate));
    memset(f, -1, sizeof(f));
    for(int i = 1; i <= k; i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        coordinate[x][y]=true;
    }
    int ans=0;
    for(int i = 1; i <= n; i++)
    {
        memset(vis, false, sizeof(vis));
        if(find(i))
        ans++;
    }
    printf("%d\n", ans);

    return 0;
}