POJ 2236 Wireless Network(并查集)

本文描述了在东南亚地震后,ACM团队如何重建受损的无线网络,通过修复计算机并确保它们在一定距离内可以相互通信,逐步恢复网络功能。文章详细介绍了输入输出格式,以及如何通过算法解决计算机间通信的问题。

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

题意描述:
第一行给出两个数字m和d,m代表的是有m台电脑,d代表的是电脑之间信息能传递的最大距离,第2行到m+1行给出的是m个电脑的坐标,然后给一个字母O和一个数字p,表示p这个点的电脑已经修好了,然后会给出一个字母S,和两个数字p和q,问p和q能通信成功吗?

#include<stdio.h>
#include<string.h>
#include<math.h>

int f[1100],n,d;

struct node
{
	int x,y;
	int s; 
}p[1100];

void inti ()
{
	for(int i=1;i<=n;i++)
		f[i]=i;
	return ;
}

int getf(int i)
{
	if(i==f[i])
		return i;
	
	f[i]=getf(f[i]);
	return f[i];
}

void merge(int n,int j)
{
	int t1,t2;
	t1=getf(j);
	for(int i=1;i<=n;i++)
	{
		t2=getf(i);
		if((p[i].s==0)&&((double)sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)*1.0+(p[i].y-p[j].y)*(p[i].y-p[j].y)*1.0)<=d))//判断这个点的电脑是否修好和到这个点的距离是不是小于等于d
			f[t2]=t1;	
	}
	return ;
}

int main(void)
{
	int a,b,c;
	char ch;
	while(scanf("%d%d",&n,&d)!=EOF)
	{
		inti();
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&p[i].x,&p[i].y);
			p[i].s=1;//标记此处电脑表示没有修好
		}
		while(~scanf(" %c",&ch))
		{
			
			if(ch=='O')
			{
				scanf("%d",&a);
				p[a].s=0;//释放标记,表示p这个点电脑修好了
				merge(n,a);
			} 
			 if(ch=='S')
			{
				scanf("%d%d",&c,&b);
				
				int t1=getf(c);
				int t2=getf(b); 
				
				if(t1!=t2)
					printf("FAIL\n");
				else
					printf("SUCCESS\n");
			}
		}	
	}	 
	return 0;
}
内容概要:本文为《科技类企业品牌传播白皮书》,系统阐述了新闻媒体发稿、自媒体博主种草与短视频矩阵覆盖三大核心传播策略,并结合“传声港”平台的AI工具与资源整合能力,提出适配科技企业的品牌传播解决方案。文章深入分析科技企业传播的特殊性,包括受众圈层化、技术复杂性与传播通俗性的矛盾、产品生命周期影响及2024-2025年传播新趋势,强调从“技术输出”向“价值引领”的战略升级。针对三种传播方式,分别从适用场景、操作流程、效果评估、成本效益、风险防控等方面提供详尽指南,并通过平台AI能力实现资源智能匹配、内容精准投放与全链路效果追踪,最终构建“信任—种草—曝光”三位一体的传播闭环。; 适合人群:科技类企业品牌与市场负责人、公关传播从业者、数字营销管理者及初创科技公司创始人;具备一定品牌传播基础,关注效果可量化与AI工具赋能的专业人士。; 使用场景及目标:①制定科技产品全生命周期的品牌传播策略;②优化媒体发稿、KOL合作与短视频运营的资源配置与ROI;③借助AI平台实现传播内容的精准触达、效果监测与风险控制;④提升品牌在技术可信度、用户信任与市场影响力方面的综合竞争力。; 阅读建议:建议结合传声港平台的实际工具模块(如AI选媒、达人匹配、数据驾驶舱)进行对照阅读,重点关注各阶段的标准化流程与数据指标基准,将理论策略与平台实操深度融合,推动品牌传播从经验驱动转向数据与工具双驱动。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值