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

思路：

        這就是一個變形的並查集，無非是加了一個距離條件，只需要判斷就好了，沒修一臺電腦，就講這個電腦和已經修好的電腦進行並查集運算，找到他的boos，然後在判斷最後輸入的兩個數的boos 是不是一個就可以了

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define inf 99999999
using namespace std;
int a[1300],tp[1300],tpx;
struct node
{
    int x;
    int y;
}b[1300];
int  judge(node a, node b,int m)
{
    double s= sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
    if(s<=m) return 1;
    else  return 0;
}
int get_boss(int x)
{
    if(x==a[x]) return x;
    else
    {
        a[x]=get_boss(a[x]);
        return a[x];
    }
}
void mergy(int x, int y)
{
    int t1,t2;
    t1=get_boss(x);
    t2=get_boss(y);
    if(t1!=t2)
    {
        a[t2]=t1;
    }
    return ;
}
int main ()
{
    int n,m,x,y;
    scanf("%d %d",&n,&m);
    //init
    for(int i=1;i<=n;i++)
    a[i]=i;
    for(int i=1; i<=n; i++)
    {
        scanf("%d %d",&b[i].x, &b[i].y);
    }
    char s;
    tpx=0;
    while (cin>>s)
    {
        if(s=='O')
        {
            scanf("%d",&x);
            for(int i=0;i<tpx;i++) // 盲搜，會重複，影響效率
            {
                if(judge(b[tp[i]],b[x],m))
                {
                    mergy(tp[i],x);
                }
            }
            tp[tpx]=x;
            tpx++;
        }
        else if(s=='S')
        {
            scanf("%d %d",&x,&y);
            x=get_boss(x);
            y=get_boss(y);
            if(x==y) printf("SUCCESS\n");
            else  printf("FAIL\n");
        }
    }
    return 0;
}