POJ 3041 Asteroids 求最大匹配数

最新推荐文章于 2021-03-29 09:55:40 发布

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最新推荐文章于 2021-03-29 09:55:40 发布

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本文介绍了一个通过最大匹配算法解决网格中障碍消除问题的方法。利用行和列作为顶点集，将障碍物位置转换为连接行与列的边。通过求解最大匹配数来找到消除所有障碍所需的最少射击次数。

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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12199		Accepted: 6641

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).

Source

USACO 2005 November Gold

把方阵的行和列看作两个顶点集v1和v2.把障碍物的x和y看成连接x和y的坐标。然后求x和y的最大匹配数就行。

#include<stdio.h>
#include<string.h>
using namespace std;
int g[507][507],link[507];
int n,m;
bool vis[507];
bool find(int i)
{
    for(int j=1;j<=n;j++)
    {
        if(g[i][j]&&!vis[j])
        {
            vis[j]=true;
            if(!link[j]||find(link[j]))
            {
                link[j]=i;
                return true;
            }

        }
    }
    return false;
}
int main()
{
    scanf("%d%d",&n,&m);
    int a,b;
    memset(g,0,sizeof(g));
    memset(link,0,sizeof(link));
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        g[a][b]=1;
    }
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof(vis));
        if(find(i))
        ans++;
    }
    printf("%d\n",ans);
    return 0;
}