CodeForces 218C 并查集

最新推荐文章于 2021-02-26 18:52:00 发布

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本文介绍了一道编程题的解决方法，题目要求通过创建最少数量的额外雪堆，使得角色可以在所有现有雪堆间通过特定的滑行方式相互到达。使用并查集算法来确定最少需要增加的雪堆数量。

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http://acm.hust.edu.cn/vjudge/contest/view.action?cid=42071#problem/C

Description

Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.

We assume that Bajtek can only heap up snow drifts at integer coordinates.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 100) — the number of snow drifts. Each of the following n lines contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ 1000) — the coordinates of the i-th snow drift.

Note that the north direction coinсides with the direction of Oy axis, so the east direction coinсides with the direction of the Ox axis. All snow drift's locations are distinct.

Output

Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.

Sample Input

Input

2
2 1
1 2

Output

Input

2
2 1
4 1

Output

并查集题目：任意两点在一行或者在一列就合并为同一棵树，最后清点树的数目n。而n-1就是答案！

代码如下：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int n;
int ab[1005];
struct note
{
    int x,y;
};
note a[1005];
void isis()
{
    for(int i=0;i<=n;i++)
       ab[i]=i;
}
int find(int x)//查找根节点
{
    if(ab[x]==x)
       return x;
    return ab[x]=find(ab[x]);
}
void  un(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x!=y)
      ab[x]=y;
}
int main()
{
    while(~scanf("%d",&n))
    {
        isis();
        for(int i=0;i<n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        for(int i=0;i<n;i++)
           for(int j=0;j<n;j++)
           {
               if(i==j)//可以没有这个if语句
                  continue;
              if(a[j].x==a[i].x||a[j].y==a[i].y)//判断是否是同一行或者同一列
              {
                  un(i,j);
              }
           }
        int count=0;
        for(int i=0;i<n;i++)
            if(ab[i]==i)//根节点是它本身一定代表着一棵树
               count++;
        printf("%d\n",count-1);
    }
    return 0;
}