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:
    前端 HTML
    python3内置函数
    内置函数的随机验证码
    线程、进程以及协程,上下文管理器
    线程池的定义方法
    python_控制台输出带颜色的文字方法
    while 循环 continue break 用法例子
    JVM 基础知识
    ios 设置状态栏文本颜色为白色
    ios 常用第三方库要加的framework,ARC的设置
  • 原文地址:https://www.cnblogs.com/ZhaoxiCheung/p/5907405.html
Copyright © 2011-2022 走看看