zoukankan      html  css  js  c++  java
  • poj-2236-Wireless Network

    Wireless Network

    Time Limit: 10000MS   Memory Limit: 65536K
    Total Submissions: 24155   Accepted: 10045

    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

    题目大意:将修好的电脑连通,随后询问俩电脑是否能接通

    题目链接:http://poj.org/problem?id=2236

     

    代码:

    #include <iostream>
    #include <cstring>
    using namespace std;
    struct ss
    {
        int x,y;
    }s[1010];
    int map[1010][1010];
    int vis[1010];
    int f[1010];
    int find(int n)
    {
        if(n!=f[n])
        f[n]=find(f[n]);
        return f[n];
    }
    int main()
    { int n,d;
        cin>>n>>d;
        for(int i=1;i<=n;i++)
        {
            cin>>s[i].x>>s[i].y;
            map[i][i]=0;
            for(int j=i-1;j>0;j--)
            map[i][j]=map[j][i]=(s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
        }
        for(int i=1;i<=n;i++)
        f[i]=i;
        memset(vis,0,sizeof(vis));
        char a[10];
        int b,c;
        while(cin>>a)
        {
            if(a[0]=='O')
            {
                cin>>b;
                vis[b]=1;
                for(int i=1;i<=n;i++)
                {
                    if(map[b][i]<=d*d&&i!=b&&vis[i])
                    {
                        int a1=find(b);
                        int b1=find(i);
                        if(a1!=b1);
                        f[b1]=a1;
                    }
                }
            }
            else
            {
                cin>>b>>c;
                if(vis[b]&&vis[c]&&find(b)==find(c))
                cout<<"SUCCESS"<<endl;
                else
                cout<<"FAIL"<<endl;
            }
        }
    
        return 0;
    }
  • 相关阅读:
    135.002 智能合约设计-——多员工薪酬系统
    131.007 Unsupervised Learning
    131.006 Unsupervised Learning
    131.005 Unsupervised Learning
    135.001 智能合约设计-——单员工薪酬系统
    131.004 监督学习项目 | 为CharityML寻找捐献者
    131.003 数据预处理之Dummy Variable & One-Hot Encoding
    Chromebook 阿里云ECS 配置 jupyter Notebook
    Python之实现迭代器协议
    使用生成器创建新的迭代模式
  • 原文地址:https://www.cnblogs.com/llsq/p/5881188.html
Copyright © 2011-2022 走看看