Codeforces---Ice Skating

题目链接： http://codeforces.com/contest/218/problem/C  

Ice Skating

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 xi and yi (1 ≤ xi, yi ≤ 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 14 1

output

0

算法： 并查集。

这个题目我是把能够彼此到达（中间可以通过其他点）的那些点归纳到一个集合中，最后统计集合的数目，需要新加的点就是集合的数目减1。

代码：

#include <iostream>
#include <string>
#include <algorithm>
using name
4000
space std;

const int N = 101;

int father
;
int n;

void init() //赋初值
{
for(int i=0;i<n;i++)
{
father[i] = i;
}
}

int find(int x) //查找
{
if(x!=father[x])
{
father[x] = find(father[x]);
}
return father[x];
}

void merge(int x, int y)
{
x = find(x);
y = find(y);
if(x == y)
{
return;
}
father[x] = y;
}

int count() //计数
{
int ans = 0;
for(int i=0; i<n; i++ )
{
if(find(i) == i)
{
ans++;
}
}
return ans;
}

struct Point
{
int x,y;
bool can_reach(Point b)
{
if(x == b.x || y == b.y)
{
return true;
}
return false;
}
}p
;

int main()
{
cin>>n;
init();
for(int i=0;i<n;i++)
{
cin>>p[i].x>>p[i].y;
}
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(p[i].can_reach(p[j]))
{
merge(i,j);
}
}
}
cout<<count()-1<<endl;
return 0;
}