Wireless Network
Time Limit: 10000MS   Memory Limit: 65536K
Total Submissions: 20724   Accepted: 8711

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
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <cmath>
using namespace std;
#define Max 1005
int n,d;
int per[Max];
bool vis[Max];
struct point
{
int x,y;
}e[];
void init()
{
for(int i=;i<=n;i++)
per[i]=i;
return;
}
int find(int x)
{
if(x==per[x])
return x;
int tem,root=x,t=x;
while(root!=per[root]) root=per[root];
while(t!=per[t])
{
tem=per[t];
per[t]=root;
t=tem;
}
}
bool dis(int q,int p)
{
point a=e[q],b=e[p];
int f=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
return d*d>=f;
}
int unite(int a,int b)
{
a=find(a);
b=find(b);
if(a!=b)
per[a]=b;
return ;
}
int main()
{
int i,j;
int v,u,p;
char ch;
freopen("in.txt","r",stdin);
scanf("%d%d",&n,&d);
init();
memset(vis,,sizeof(vis));
for(i=;i<=n;i++)
scanf("%d%d",&e[i].x,&e[i].y);
getchar();
while(scanf("%c",&ch)!=EOF)
{
if(ch=='S')
{
scanf("%d%d",&u,&v);
if(find(u)==find(v))
printf("SUCCESS\n");
else
printf("FAIL\n");
}
else if(ch=='O')
{
scanf("%d",&u);
vis[u]=;
for(i=;i<=n;i++)
if(vis[i]&&i!=u&&dis(i,u))
unite(u,i);
}
getchar();
}
return ;
}

Wireless Network(POJ 2236)的更多相关文章

  1. Day5 - B - Wireless Network POJ - 2236

    An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wi ...

  2. (并查集) Wireless Network --POJ --2236

    链接: http://poj.org/problem?id=2236 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82830#probl ...

  3. DisJSet:Wireless Network(POJ 2236)

      无线电网络 题目大意:就是地震后,所有的电脑都坏了,现在可以修复,而且要重新连成一个网络,两台电脑之间最大连接距离为D,两台电脑可以有中继电脑,按O修复电脑,按S测试两台电脑是否有链接,如果有就输 ...

  4. Wireless Network POJ - 2236 (并查集)

    #include<iostream> #include<vector> #include<string> #include<cmath> #includ ...

  5. A - Wireless Network POJ - 2236

    题目大意:有n台坏掉的电脑,给出每台电脑的坐标,然后每次询问输入0(字符) x,表示电脑x恢复正常,输入S x y 询问x和y是否可以联网.只要是x和y的距离小于距离d,那么就可以联网,如果有个中介c ...

  6. A - Wireless Network POJ - 2236-kuangbin带你飞

    A - Wireless Network POJ - 2236 Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 50348 ...

  7. POJ 2236 Wireless Network ||POJ 1703 Find them, Catch them 并查集

    POJ 2236 Wireless Network http://poj.org/problem?id=2236 题目大意: 给你N台损坏的电脑坐标,这些电脑只能与不超过距离d的电脑通信,但如果x和y ...

  8. [并查集] POJ 2236 Wireless Network

    Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 25022   Accepted: 103 ...

  9. poj 2236:Wireless Network(并查集,提高题)

    Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 16065   Accepted: 677 ...

随机推荐

  1. CActiveForm提示中文化

    一般情况下,yii框架的CActiveForm组建的提示是英文的,把它改成中文提示 首先在main.php配置文件中,设置 'language'=>'zh_cn'; 这样将会使得cannot b ...

  2. 单片机 认识HEX文件

    看过几篇常用指令的用法后,我们换换口味,介绍一下Intel 原厂所公布的HEX文件标准格式,相信经过本文的介绍,一定可以让您对8051的操作有更进一步的认识.以下是一个程序经编译器编译后所得到的HEX ...

  3. 通用Makefile

    本文推荐了一个用于对 C/C++ 程序进行编译和连接以产生可执行程序的通用 Makefile. 在使用 Makefile 之前,只需对它进行一些简单的设置即可:而且一经设置,即使以后对源程序文件有所增 ...

  4. Merge Sorted Array 解答

    Question Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array ...

  5. Hdu3812-Sea Sky(深搜+剪枝)

    Sea and Sky are the most favorite things of iSea, even when he was a small child.  Suzi once wrote: ...

  6. Colorful Lecture Note(手工栈)

    题目1 : Colorful Lecture Note 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 Little Hi is writing an algorithm ...

  7. python海明距离 - 5IVI4I_I_60Y的日志 - 网易博客

    python海明距离 - 5IVI4I_I_60Y的日志 - 网易博客 python海明距离   2009-10-01 09:50:41|  分类: Python |  标签: |举报 |字号大中小  ...

  8. xsank的快餐 » Python simhash算法解决字符串相似问题

    xsank的快餐 » Python simhash算法解决字符串相似问题 Python simhash算法解决字符串相似问题

  9. ps查看内存占用排序

    ps -eo rss,pmem,pcpu,vsize,args | sort -k 1 -r -n | less 解析一下: ps 都知道,是linux,unix显示进程信息的, -e 是显示所有进程 ...

  10. Android基础&进阶

    http://blog.csdn.net/liuhe688/article/details/9494411