题目：

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

题意是有N台电脑坏了，需要维修，而修好的电脑之间如果距离小于等于D,它们就可以连通，而两台电脑之间可以借助第三台电脑连接。O代表维修电脑，S代表检测两台电脑是否连通，是就输出SUCCESS，否则就输出FAIL.

由于每台电脑都有坐标，因此可以算出两两之间的距离。判断是否连接就得先知道这两台电脑是否都已经修好，可以用一个数组存储修好的电脑的编号。

由于要判断是否连通，就等价于判断两台电脑是否属于同一个集合，因此可以用并查集，先将每一台修好的电脑与之前已修好的电脑的距离求出，符合的话就加入对应的集合。
最后判断两台电脑是否属于同一个集合，是就能连通，否则就无法连通。

参考代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define N 1005
using namespace std;
struct point {
    int x,y;
}p[N];
int repaired[N];
int par[N],ranki[N];
void init(int n) {
    for (int i = 1;i <= n;++i) {
        par[i] = i;
        ranki[i] = 0;
    }
}
double distp(int x1,int x2,int y1,int y2) {
    double d1;
    d1 = sqrt((double)((x1-x2)*(x1-x2)) + (double)((y1-y2)*(y1-y2)));
    return d1;
}
int find(int x) {
    if (par[x] == x) return x;
    else return par[x] = find(par[x]);  
}
void unite(int x,int y) {
    int fx = find(x);
    int fy = find(y);
    if (fx == fy) return;
    if (ranki[fx] > ranki[fy]) par[fy] = fx;
    else {
        par[fx] = fy;
        if (ranki[fx] == ranki[fy]) ranki[fy]++;
    }
}
int main() {
    int n,d;
    while (cin >> n >> d) {
        memset(p,0,sizeof(p));
        for (int i = 1;i <= n;++i) {
            cin >> p[i].x >> p[i].y;
        }
        init(n);
        char op;
        int num1;
        int nump,numq;
        int cnt = 1;
        while (cin >> op) {
            if (op == 'O') {
                cin >> num1;
                for (int i = 1;i <= cnt;++i) {
                    double dis = distp(p[repaired[i]].x,p[num1].x,p[repaired[i]].y,p[num1].y);
                    if (dis <= (double)(d)) {
                        unite(repaired[i],num1);
                    }
                }
                repaired[cnt++] = num1;
            }
            else if (op == 'S') {
                cin >> nump >> numq;
                nump = find(nump);
                numq = find(numq);
                if (nump == numq) cout << "SUCCESS" << endl;
                else cout << "FAIL" << endl;    
            }
        }
    }
    return 0;
}