kuangbin专题五A题- Wireless Network POJ - 2236

 

A - Wireless Network

POJ - 2236

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：刚开始 有 N 台坏掉的计算机 ， 然后 距离相差不超过 d 的计算机（已经修复）之间可以 联系 。
　　　计算机 从 1 到 N 编号 ， 之后 输入 N 台计算机 的位置 ， 
　　　然后 有多组操作 O x （表示 编号为 x 的电脑被修复） ，S x y （表示 查询 电脑 x 和 电脑 y 之间能不能联系（包括直接联系和间接联系））
　　　可以联系 输出 SUCCESS ， 不能联系 输出 FAIL 
思路：并查集 

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

#define maxn 1010 
bool flag[maxn] ;
int father[maxn] ; 

vector<int> vv[maxn] ; 
int n , d ; 

struct node {
    int x , y ;
} post[maxn] ; 

double fun(int x){
    return x*x*1.0 ; 
}

void init(){
    memset(vv , 0 , sizeof(vv)) ; 

    for(int i=1 ; i<=n ; i++){
        father[i] = i ; //初始化 每台电脑为一个集合
        flag[i] = false ;// 所有电脑 初始状态为损坏 

        for(int j =1 ; j<=n ; j++){
            if(i==j){
                continue ; 
            }
            //为 i 电脑 链接 所有 可连接距离内的电脑
            double dis = sqrt(fun(post[i].x-post[j].x)+fun(post[i].y-post[j].y)) ; 
            if(dis<=d){
                vv[i].push_back(j) ; 
            }
        }
    }
}
//查询 x 所在集合的代表元素的编号
//同时 理顺集合内元素的关系（为刚合并的两个集合找出统一的代表）
int find(int x){
    if(x!=father[x])
        father[x] = find(father[x]) ; 
    return father[x] ; 
}

void Union_set(int x , int y){
    int rootx = find(x) ; 
    int rooty = find(y) ; 

    if(rootx != rooty){
        father[rooty] = rootx ; // 将 第二个集合的带表元素 连接到第一个集合 
    }
}



int main(){

    while(~scanf("%d %d" , &n , &d)){
        for(int i=1 ; i<=n ; i++){
            scanf("%d%d" , &post[i].x , &post[i].y) ; 
        }

        init() ; 
        char str[3] ; 
        int x , y ; 

        while(~scanf(" %s" , str)){
            //scanf(" %s" , str) ; 

            if(str[0] == 'O'){
                scanf("%d" , &x) ; 
                flag[x] = true ; 
                for(int j=0 ; j<vv[x].size() ; j++){
                    int k = vv[x][j] ; 
                    if(flag[k]){//编号为 k 的电脑已经被修复 
                        Union_set(x , k ) ; 
                    }
                }
            }else if(str[0] =='S'){
                scanf("%d %d" , &x , &y) ; 
                int rootx = find(x) ; 
                int rooty = find(y) ; 
                if(rootx == rooty){
                    printf("SUCCESS\n") ; 
                } else {
                    printf("FAIL\n") ; 
                }
            }
        }
        
    }
    return 0 ; 
}