zoukankan      html  css  js  c++  java
  • POJ 2236 Wireless Network(并查集)

    传送门

     Wireless Network

    Time Limit: 10000MS   Memory Limit: 65536K
    Total Submissions: 24513   Accepted: 10227

     

    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
    

    思路

    并查集模板题,奇怪的是用G++跑了3000MS多,C++则是1000MS多
     
    #include<stdio.h>
    #include<string.h>
    const int maxn = 1005;
    struct Node{
    	int x,y;
    }node[maxn];
    int N,d,fa[maxn];
    bool repair[maxn];
    
    int find(int x)
    {
    	int r = x;
    	while (r != fa[r])	r = fa[r];
    	int i = x,j;
    	while (i != r)
    	{
    		j = fa[i];
    		fa[i] = r;
    		i = j;
    	}
    	return r;
    }
    
    void unite(int x,int y)
    {
    	x = find(x),y = find(y);
    	if (x != y)	fa[x] = y;
    }
    
    bool dist(struct Node xx,struct Node yy)
    {
    	return	((xx.x-yy.x)*(xx.x-yy.x)+(xx.y-yy.y)*(xx.y-yy.y)) <= d*d;
    }
    
    int main()
    {
    	int index,index1,index2;
    	char op[5];
    	memset(repair,false,sizeof(repair));
    	scanf("%d%d",&N,&d);
    	for (int i = 1;i <= N;i++)
    	{
    		fa[i] = i;
    		scanf("%d%d",&node[i].x,&node[i].y);
    	}
    	while (~scanf("%s",op))
    	{
    		if (op[0] == 'O')
    		{
    			scanf("%d",&index);
    			repair[index] = true;
    			for (int i = 1;i <= N;i++)
    			{
    				if (i != index && repair[i] && dist(node[i],node[index]))	unite(i,index);
    			}
    		}
    		else if (op[0] == 'S')
    		{
    			scanf("%d%d",&index1,&index2);
    			index1 = find(index1),index2 = find(index2);
    			if (index1 == index2)	printf("SUCCESS
    ");
    			else	printf("FAIL
    ");
    		}
    	}
    	return 0;
    }
  • 相关阅读:
    非域,非匿名用户访问远程企业服务的详细步骤
    调用远程的企业服务的安全问题
    未能加载文件或程序集“****”或它的某一个依赖项的一种情况
    XAMPP使用非80端口的安装配置修改
    Lucene 的存储结构概述
    .NET Framework 4.0 SDK的安装
    lucene 文件存储相关的几个类
    ASP.NET 状态服务 及 session丢失问题解决方案总结
    不安装.net framework框架运行.Net 程序的方法<收藏>
    net面试题集及答案
  • 原文地址:https://www.cnblogs.com/ZhaoxiCheung/p/5907405.html
Copyright © 2011-2022 走看看