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.
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 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.
2 2 1 1 2
1
2 2 1 4 1
0
【题意】
给出n个点的横、纵坐标（在一条直线上的可以相互到达），问至少再加几个点可以使得所有的点之间可以通过上下左右走互相到达，（拐点上必须有“点”）
【分析】
若两个点横坐标或者纵坐标相同，两点间连一条边 通过dfs统计连通块的个数
假设处理完后有t个缩点，只需t-1条边就可将其变成一个缩点。故答案为t-1.
当然并查集也是可以的，但dfs更轻便
【代码】
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef pair<int,int> pir;
const int N=105;
int n,ans;pir e[N];bool vis[N],g[N][N];
void dfs(int x){
vis[x]=1;
for(int j=1;j<=n;j++){
if(!vis[j]&&g[x][j]){
dfs(j);
}
}
}
#define x first
#define y second
inline void Solve(){
for(int i=1;i<=n;i++){
for(int j=1;j<i;j++){
if(e[i].x==e[j].x||e[i].y==e[j].y){
g[i][j]=g[j][i]=1;
}
}
}
for(int i=1;i<=n;i++) if(!vis[i]) dfs(i),ans++;
printf("%d",ans-1);
}
inline void Init(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d%d",&e[i].x,&e[i].y);
}
int main(){
Init();
Solve();
return 0;
}