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

    Wireless Network
    Time Limit: 10000MS   Memory Limit: 65536K
    Total Submissions: 20724   Accepted: 8711

    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
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <set>
     6 #include <cmath>
     7 using namespace std;
     8 #define Max 1005
     9 int n,d;
    10 int per[Max];
    11 bool vis[Max];
    12 struct point
    13 {
    14     int x,y;    
    15 }e[1002];
    16 void init()
    17 {
    18     for(int i=0;i<=n;i++)
    19         per[i]=i;
    20     return;
    21 }
    22 int find(int x)
    23 {
    24     if(x==per[x])
    25         return x;
    26     int tem,root=x,t=x;
    27     while(root!=per[root])    root=per[root];
    28     while(t!=per[t])
    29     {
    30         tem=per[t];
    31         per[t]=root;
    32         t=tem;
    33     }
    34 }
    35 bool dis(int q,int p)
    36 {
    37     point a=e[q],b=e[p];
    38     int f=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
    39     return d*d>=f;
    40 }
    41 int unite(int a,int b)
    42 {
    43     a=find(a);
    44     b=find(b);
    45     if(a!=b)
    46         per[a]=b;
    47     return 0;
    48 }
    49 int main()
    50 {
    51     int i,j;
    52     int v,u,p;
    53     char ch;
    54     freopen("in.txt","r",stdin);
    55     scanf("%d%d",&n,&d);
    56     init();
    57     memset(vis,0,sizeof(vis));
    58     for(i=1;i<=n;i++)
    59         scanf("%d%d",&e[i].x,&e[i].y);
    60     getchar();
    61     while(scanf("%c",&ch)!=EOF)
    62     {
    63         if(ch=='S')
    64         {
    65             scanf("%d%d",&u,&v);
    66             if(find(u)==find(v))
    67                 printf("SUCCESS
    ");
    68             else
    69                 printf("FAIL
    ");
    70         }
    71         else if(ch=='O')
    72         {
    73             scanf("%d",&u);
    74             vis[u]=1;
    75             for(i=1;i<=n;i++)
    76                 if(vis[i]&&i!=u&&dis(i,u))
    77                     unite(u,i);
    78         }
    79         getchar();
    80     }
    81     return 0;
    82 }
  • 相关阅读:
    最短路径(Dijkstra算法)
    图的优先级搜索
    图的遍历(搜索)

    树(二叉树)
    TF-池化函数 tf.nn.max_pool 的介绍
    TF-卷积函数 tf.nn.conv2d 介绍
    TF-图像的深度和通道的概念(转)
    MongoDB-MongoDB重装系统后恢复
    MYSQL-重做系统恢复MYSQL过程
  • 原文地址:https://www.cnblogs.com/a1225234/p/5181208.html
Copyright © 2011-2022 走看看