zoukankan      html  css  js  c++  java
  • poj-2236 wireless network(并查集)

    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:修复和测试,然后输出结果
    题解:并查集的模版题,直接看代码吧
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<sstream>
    #include<cmath>
    #include<cstdlib>
    #include<queue>
    using namespace std;
    #define PI 3.14159265358979323846264338327950
    
    int father[20010],num;
    int p[20010]={0},n;
    
    struct point
    {
        double x;
        double y;
    }pt[1005];
    
    
    int find(int x)
    {
        if(x==father[x])
            return x;
        return father[x]=find(father[x]);
    }
    
    double dis(point a,point b)
    {
        return sqrt((a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y)*1.0);
    }
    
    void combine(int a,int b)
    {
        int fa=find(a);
        int fb=find(b);
        if(fa!=fb)
            father[fa]=fb;
    }
    
    int main()
    {
        int i,x,y;
        double d;
        char control[2];
        scanf("%d %lf",&n,&d);
        
      
        for(int i=0;i<1005;i++)
            father[i]=i;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf %lf",&pt[i].x,&pt[i].y);
        }
        while(scanf("%s",control)!=EOF)
        {
            if(control[0]=='O')
            {
                scanf("%d",&x);
                p[x]=1;
                for(i=1;i<=n;i++)
                {
                    if(dis(pt[i],pt[x])<=d&&p[i]==1)
                        combine(i,x);
                }
                
            }
            if(control[0]=='S')
            {
                scanf("%d%d",&x,&y);
                if(find(x)==find(y))
                    printf("SUCCESS
    ");
                else
                    printf("FAIL
    ");
            }
        }
        
        
    }
  • 相关阅读:
    公用导航栏的根据url控制选中导航js
    页面切换出动晃动解决
    redis五大数据类型
    redis简介
    Linux安装redis
    各种锁的理解
    原子引用
    理解CAS
    彻底玩转单例模式
    Volatile
  • 原文地址:https://www.cnblogs.com/smallhester/p/9498728.html
Copyright © 2011-2022 走看看