POJ 2236  Wireless Network

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 10451		Accepted: 4431

Description

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired one by one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded as the intermediary of the communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.

In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.

Input

The first line contains two integers N and d (1 <= N <= 1001, 0 <= d <= 20000). Here N is the number of computers, which are numbered from 1 to N, and D is the maximum distance two computers can communicate directly. In the next N lines, each contains two integers xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following two formats:
1. "O p" (1 <= p <= N), which means repairing computer p.
2. "S p q" (1 <= p, q <= N), which means testing whether computer p and q can communicate.

The input will not exceed 300000 lines.

Output

For each Testing operation, print "SUCCESS" if the two computers can communicate, or "FAIL" if not.

Sample Input

4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4

Sample Output

FAIL
SUCCESS

Source

POJ Monthly,HQM
好激动，看了别人的解释后直接AC！
引用大神的解释：
本题的题意 是判断俩点之间的连通性，不过是带受限制条件的判断题，主要把握好这个判断就可以了，用一个数组记录该点是否可行，每次接收一个点就对整个数组经行一次扫 描 把在限制条件内的点都并入集合 ，然后判断就直接用Find(x)==Find(y)判断俩个点是否连通就可以了
代码：
#include<stdio.h>
#include<math.h>
struct point{
      int x,y;
      int index;
}num[1008];
int father(int x)
{
      if(num[x].index==x)
            return x;
      num[x].index=father(num[x].index);
      return num[x].index;
}
double distance(int a,int b)
{
      return sqrt((num[a].x-num[b].x)*(num[a].x-num[b].x)+(num[a].y-num[b].y)*(num[a].y-num[b].y));
}
int main()
{
      int n,d,i,p,q,ra,rb;
      int flag[1008]={0};
      char s;
      scanf("%d%d",&n,&d);
      for(i=1;i<=n;i++)
      {
            scanf("%d%d",&num[i].x,&num[i].y);
            num[i].index=i;
      }
      getchar();
      while(scanf("%c %d",&s,&p)!=EOF)
      {
            if(s=='O')
            {
                  flag[p]=1;
                  for(i=1;i<=n;i++)
                        if(distance(p,i)<=d+0.0000001&&flag[i]&&p!=i)
                        {
                              ra=father(p);
                              rb=father(i);
                              if(ra>rb)
                                    num[ra].index=rb;
                              else num[rb].index=ra;
                        }
            }
            if(s=='S'){
                  scanf("%d",&q);
                  if(flag[p]&&flag[q]){
                  ra=father(p);
                  rb=father(q);
                  if(ra==rb)printf("SUCCESS\n");
                  else printf("FAIL\n");}
                  else printf("FAIL\n");
            }
            getchar();
      }
      return 0;
}