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 这道题可以说是最简单的并查集,直接套模板其实就OK了
现附上AC代码:

#include<iostream>
#include<cstdio>
using namespace std;
struct node{
int a,plug;int x,y;
}com[1010]; //结构体变量存储a:并查集的标准量等于自身;plug:标志量,若为1则电脑修好,为0未修好;(x,y):存储电脑的坐标
int N,d;
int find(int x); //找到祖先
void unite(int i,int p); //合并并查集
int main(){
cin>>N>>d;
char w; //这个变量最扯了,刚开始没有用到这个变量,用的是:fflush(stdin),结果wrong answer,改为%c,&w,吃掉空格换行符就可以,郁闷了。
for(int i=1;i<=N;i++)
scanf("%d%c%d%c",&com[i].x,&w,&com[i].y,&w),com[i].a=i,com[i].plug=0;//电脑坐标,以及初始化
char c;
int p,x,y;
//fflush(stdin);
while(~scanf("%c%c",&c,&w)){
if(c=='O') {
scanf("%d%c",&p,&w);
for(int i=1;i<=N;i++)
if(com[i].plug&&(com[i].x-com[p].x)*(com[i].x-com[p].x)+(com[i].y-com[p].y)*(com[i].y-com[p].y)<=d*d) unite(i,p); //满足条件就合并
com[p].plug=1;
}
else {
scanf("%d%c%d%c",&x,&w,&y,&w);
if(find(x)==find(y)) printf("SUCCESS\n"); //查询是否可以相连
else printf("FAIL\n");
}
//fflush(stdin);
}
return 0;
}
int find(int x){
if(com[x].a==x) return x;
else return com[x].a=find(com[x].a);
}
void unite(int i,int p){
i=find(i),p=find(p);
com[i].a=p;
}

poj2236Wireless Network的更多相关文章

  1. poj-2236-Wireless Network

    Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 24155   Accepted: 100 ...

  2. 【并查集】POJ2236-Wireless Network

    [题目大意] 已知每一台电脑只能与它距离为d的电脑相连通,但是两台电脑间可以以第三台作为媒介连接.现在电脑全被损坏.每次可以进行两个操作中的一个,或是修好一台电脑,或是查询两台电脑是否连通. [思路] ...

  3. Recurrent Neural Network系列1--RNN(循环神经网络)概述

    作者:zhbzz2007 出处:http://www.cnblogs.com/zhbzz2007 欢迎转载,也请保留这段声明.谢谢! 本文翻译自 RECURRENT NEURAL NETWORKS T ...

  4. 创建 OVS flat network - 每天5分钟玩转 OpenStack(134)

    上一节完成了 flat 的配置工作,今天创建 OVS flat network.Admin -> Networks,点击 "Create Network" 按钮. 显示创建页 ...

  5. 在 ML2 中配置 OVS flat network - 每天5分钟玩转 OpenStack(133)

    前面讨论了 OVS local network,今天开始学习 flat network. flat network 是不带 tag 的网络,宿主机的物理网卡通过网桥与 flat network 连接, ...

  6. OVS local network 连通性分析 - 每天5分钟玩转 OpenStack(132)

    前面已经创建了两个 OVS local network,今天详细分析它们之间的连通性. launch 新的 instance "cirros-vm3",网络选择 second_lo ...

  7. 再部署一个 instance 和 Local Network - 每天5分钟玩转 OpenStack(131)

    上一节部署了 cirros-vm1 到 first_local_net,今天我们将再部署 cirros-vm2 到同一网络,并创建 second_local_net. 连接第二个 instance 到 ...

  8. 创建 OVS Local Network - 每天5分钟玩转 OpenStack(129)

    上一节我们完成了 OVS 的准备工作,本节从最基础的 local network 开始学习.local network 不会与宿主机的任何物理网卡连接,流量只被限制在宿主机内,同时也不关联任何的 VL ...

  9. Configure a bridged network interface for KVM using RHEL 5.4 or later?

    environment Red Hat Enterprise Linux 5.4 or later Red Hat Enterprise Linux 6.0 or later KVM virtual ...

随机推荐

  1. [SP3267]DQUERY - D query

    题目传送门 维护一个区间内不同数的个数,最直观的想法是直接排序后用树状数组维护即可.但是我们发现n只有3e4,于是我们想到了可以拿一个$O(n\sqrt{n})$的莫队维护.关于莫队算法如果有不知道的 ...

  2. C Yuhao and a Parenthesis

    Yuhao and a Parenthesis time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  3. redis-Nosql

    Nosql: CAP:C(Consistency):强一致性.A(Availability):可用性.P(Partitio Tolerance):分区容错性 CAP 理论的核心是: 一个分布式系统,不 ...

  4. TensorFlow——实现线性回归算法

    import tensorflow as tf import numpy as np import matplotlib.pyplot as plt #使用numpy生成200个随机点 x_data= ...

  5. UNIX网络编程总结三

    套接口结构 IPv4套接口地址结构: 1 2 3 4 5 6 7 struct sockaddr_in{/*16字节*/ uint8_t sin_len;  /*结构体长度,8位*/ sa_famil ...

  6. 高精乘(fft板子

    哇..fft的原理真的是不太好懂,看了好久许多细节还是不太清楚,但感觉本质就是用了单位根的性质. https://www.luogu.org/problem/P1919 #include<cst ...

  7. C++使用静态类成员时出现的一个问题

    开发环境 Qt Creator 4.8.2 编译器 MinGw 32-bit 在类中定义了一个static data member class Triangular{ public: static b ...

  8. j函数 判断以 什么开头

    1.str.charAt(index) 返回字符串中指定位置的字符. str 是字符串  我们要将获得的数据 转化为字符串 var code = res.statusCode.toString(); ...

  9. bzoj3772 精神污染 dfs 序+主席树

    题目传送门 https://lydsy.com/JudgeOnline/problem.php?id=3772 题解 很简单的一道题目. 上午研究一个题目的时候发现了这个题目是一个弱化版,所以来写了一 ...

  10. 032:DTL常用过滤器(1)

    为什么需要过滤器: 在DTL中,不支持函数的调用形式‘()’,因此不能给函数传递参数,这将有很大的局限性:而过滤器其实就是一个函数,可以对需要处理的参数进行处理,并且还可以额外接受一个参数(也就是说: ...