poj 3041 Asteroids（最小顶点覆盖，二分图）

最新推荐文章于 2022-08-16 21:41:12 发布

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探讨了一种算法，用于解决在N*N网格中消除所有小行星所需的最小射击次数问题。通过将行视为X集合，列视为Y集合，将问题转化为最小顶点覆盖问题，并使用匈牙利算法求解。

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20107		Accepted: 10905

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

Sample Output

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).

题意：n*n的矩阵上面有一些x点，现在问你最少攻击几次可以把x点都消灭，每次攻击可以消灭一行或者一列

思路：行做为x集合，列作为y集合，然后问题就转换成一个最小顶点覆盖的问题了

直接跑匈牙利即可

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 550
int ma[N][N];
int n,line[N],vis[N];
int can(int t)
{
    for(int i=1; i<=n; i++)
    {
        if(!vis[i]&&ma[t][i])
        {
            vis[i]=1;
            if(line[i]==-1||can(line[i]))
            {
                line[i]=t;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int k;
    int x,y;
    while(~scanf("%d %d",&n,&k))
    {
        memset(line,-1,sizeof(line));
        memset(ma,0,sizeof(ma));
        for(int i=1; i<=k; i++)
        {
            scanf("%d %d",&x,&y);
            ma[x][y]=1;
        }
        int ans=0;
        for(int i=1; i<=n; i++)
        {
            memset(vis,0,sizeof(vis));
            if(can(i))
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}