poj-3041 Asteroids

最新推荐文章于 2019-09-27 18:03:26 发布

原创最新推荐文章于 2019-09-27 18:03:26 发布 · 345 阅读

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本文介绍了一种使用二分图匹配算法解决太空陨石场清除问题的方法。通过将问题抽象为寻找最小数量的行或列操作来消除所有陨石，实现了高效的解决方案。文章提供了一个具体的示例和C++实现代码。

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).

由于题中要求一次消去所在点的一行或一列，所以用二分图匹配一个x和y就相当于消去一行和一列。该题模板题。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF 0x3f3f3f3f

using namespace std;

int map[502][502];
int link[502];
int used[502];
int ans,n;
int DFS(int u)
{
    int i,j;
    for (int v=1;v<=n;v++)
    {
        if(!used[v]&&map[u][v])
        {
            used[v]=1;
            if(!link[v]||DFS(link[v]))
            {
                link[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
void maxmatch()
{
    memset(link,0,sizeof(link));
    for (int u=1;u<=n;u++)
    {
        memset(used,0,sizeof(used));
        if(DFS(u)) ans++;
    }
}
int main()
{
    int k,x,y;
    cin>>n>>k;
    ans=0;
    while (k--)
    {
        cin>>x>>y;
        map[x][y]=1;
    }
    maxmatch();
    printf("%d\n",ans);
}