poj 2236 Wireless Network（并查集）

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在东南亚发生地震后，ACM团队建立的无线网络遭到破坏。本文介绍了一种算法，用于在网络修复过程中判断任意两台计算机是否能通过直接或间接通信进行连接。算法通过维护一组可以通信的计算机集合，并在每次修复操作后更新这些集合，从而高效地回答测试操作。

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

Sample Output

FAIL
SUCCESS

题意：
每台电脑只能与距离它不超过 d 米的其它电脑直接通信。但只要两台电脑间存在一台电脑 C 既可与 A 也可与 B 通信，那么电脑 A 和电脑 B 之间就能够通信。
1. “O p” (1 <= p <= N)，表示维护电脑 p 。
2. “S p q” (1 <= p, q <= N)，表示测试电脑 p 和 q 是否能够通信。
注：只有电脑维护了才能使用

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdlib>
using namespace std;

double dis[1020][1020];//记录两点间距离
int pre[1020];//记录该点父节点
bool vis[1020] = {0};//标记此点是否存在

struct node{
    int x, y;
}a[1020];

void init(int n){
    for(int i = 1; i <= n; i++){
        pre[i] = i;
    }
}

int Find(int x){
    int r = x;
    while(pre[r] != r)
        r = pre[r];
    return r;
}

void join(int x, int y){
    int fx = Find(x);
    int fy = Find(y);
    if(fx != fy)
        pre[fx] = fy;
}

int main(){
    int n, d, i, j, x, y, t;
    char c;
    scanf("%d%d", &n, &d);
    for(i = 1; i <= n; i++){
        scanf("%d%d", &a[i].x, &a[i].y);
    }
    for(i = 1; i <= n; i++){
        for(j = 1; j <= n; j++){
            x = abs((a[i].x - a[j].x));
            y = abs((a[i].y - a[j].y));
            dis[i][j] = sqrt(x*x + y*y);
        }
    }
    init(n);
    while(scanf("%c%d", &c, &x) != EOF){
        if(c == 'O'){
            vis[x] = 1;
            for(i = 1; i <= n; i++){
                //int fx = Find(x);
                //int fy = Find(i);
               // cout << "fx = " << fx << " fy = " << fy << endl;
                if(dis[x][i] <= d && vis[i] == 1){
                    join(x, i);
                }
                //fx = Find(x);
                //fy = Find(i);
                //cout << "fx = " << fx << " fy = " << fy << endl;
            }
        }
        if(c == 'S'){
            scanf("%d", &y);
            int fx = Find(x);
            int fy = Find(y);
            //cout << "fx = " << fx << " fy = " << fy << endl;
            if(fx != fy)
                printf("FAIL\n");
            else
                printf("SUCCESS\n");
        }
    }
    return 0;
}