C - Ice Skating
Codeforces Round #134 (Div. 1)
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.
Example
Input
2
2 1
1 2
Output
1
Input
2
2 1
4 1
Output
0
題意：二維坐標系，有很多雪堆，你只能平行于x軸或y軸從一個雪堆移動到另個個雪堆處，問最少添加幾個雪堆，使你能到達全部雪堆
解法：bfs，類似油田問題，能互相到達的一些雪堆為一個塊，判斷有幾塊雪堆，減一即為要添加的雪堆個數(shù)。
代碼：
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int n;
int x[1005];
int y[1005];
bool vis[1005];
queue<int> q;
int ans=0;
void bfs()
{
while(!q.empty()){
int t=q.front();
q.pop();
vis[t]=1;
for(int i=0;i<n;i++)
if(!vis[i]&&(x[i]==x[t]||y[i]==y[t]))
q.push(i);
}
}
int main()
{
cin>>n;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
cin>>x[i]>>y[i];
for(int i=0;i<n;i++)
if(!vis[i]){
q.push(i);
bfs();
ans++;
}
cout<<ans-1<<endl;
}
