Codeforces Round 134 div 2 C题

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本题是一道关于并查集的应用题，目标是最小化新建雪堆数量，使得任意两个雪堆间都可达。通过判断已有雪堆间的相对位置关系来合并集合，最终未合并的集合数减一即为答案。

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 test(s)

input
2
2 1
1 2

output
1

input
2
2 1
4 1

output
0

这道题看似有点像数学的题目，但实际上是并查集的应用，输入的结点从1编号到n，然后初始化并查集set[i]=i,然后双重循环判断任意两点是否是一个集合的，如果两个点有x或者y相等，则两点属于一个集合。最后看有多少个集合，得到的数字减1 就是结果。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

typedef struct Node
{
       int x;
       int y;
}node;

node a[110];
int set[110];
bool vis[110];

int findSet(int x)//查找元素x所属集合 
{
    if(x==set[x]) 
       return x;
    else
       return set[x]=findSet(set[x]);
}


void unionSet(int x,int y)//x和y合并到一个集合中 
{
   int fx=findSet(x);
   int fy=findSet(y);
   set[fy]=fx;
}

int main()
{
    int n;
    scanf("%d",&n);
    int i,j;
    for(i=1;i<=110;i++)
       set[i]=i;
    for(i=1;i<=n;i++)
       scanf("%d%d",&a[i].x,&a[i].y);
    for(i=1;i<=n-1;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(i==j) continue;
            if(a[i].x==a[j].x || a[i].y==a[j].y)
            {unionSet(i,j);}
        }
    }
    int s=0;
    for(i=1;i<=n;i++)
    if(set[i]==i)  s++;//统计有多少个集合 
    printf("%d",s-1);
    //system("pause");
    return 0;
}