題目：

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

題意是有N臺電腦壞了，需要維修，而修好的電腦之間如果距離小于等于D,它們就可以連通，而兩臺電腦之間可以借助第三臺電腦連接。O代表維修電腦，S代表檢測兩臺電腦是否連通，是就輸出SUCCESS，否則就輸出FAIL.

由于每臺電腦都有坐標，因此可以算出兩兩之間的距離。判斷是否連接就得先知道這兩臺電腦是否都已經修好，可以用一個數組存儲修好的電腦的編號。

由于要判斷是否連通，就等價于判斷兩臺電腦是否屬于同一個集合，因此可以用并查集，先將每一臺修好的電腦與之前已修好的電腦的距離求出，符合的話就加入對應的集合。
最后判斷兩臺電腦是否屬于同一個集合，是就能連通，否則就無法連通。

參考代碼：

#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;
}