zoukankan      html  css  js  c++  java
  • Pet

    Pet

    Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 28   Accepted Submission(s) : 9
    Problem Description
    One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is always one distance unit.
     
    Input
    The input contains multiple test cases. Thefirst line is a positive integer T (0<t<=10), div="" <="" n="" 0.="" location="" at="" always="" room="" ji’s="" lin="" map.="" adjacent="" y="" that="" meaning="" space,="" y(0<="x,y<N)," x="" map,="" descripts="" n-1lines="" following="" trap.="" distance="" affective="" d="" school="" in="" locations="" is="" space.="" single="" a="" by="" separated="" d(0<d<n),="" and="" (0<n<="100000)" integer="" positive="" two="" has="" line="" first="" cases,="" each="" for="" cases.="" test="" of="" number="" the="">
     
    Output
    For each test case, outputin a single line the number of possible locations in the school the hamster may be found.
     
    Sample Input
    1 10 2 0 1 0 2 0 3 1 4 1 5 2 6 3 7 4 8 6 9
     
    Sample Output
    2
     
    Source
    2013 ACM/ICPC Asia Regional Online —— Warmup
     
     1 #include <stdio.h>
     2 #include <string.h>
     3 
     4 int main()
     5 {
     6     int T,i,j,head,last,sign,times,high,num[100005];
     7     scanf("%d",&T);
     8     while(T--)
     9     {
    10         memset(num,0,sizeof(num));
    11         scanf("%d%d",&times,&high);
    12         for(i=1;i<times;i++)
    13         {
    14             scanf("%d%d",&head,&last);
    15             num[last]=num[head]+1;
    16         }
    17         for(i=0,sign=0;i<=last;i++)
    18         {
    19             if(num[i]>high)
    20                 sign++;
    21         }
    22         printf("%d
    ",sign);
    23     }
    24 
    25     return 0;
    26 }
    View Code
    转载请备注:
    **************************************
    * 作者: Wurq
    * 博客: https://www.cnblogs.com/Wurq/
    * Gitee: https://gitee.com/wurq
    **************************************
  • 相关阅读:
    mongdb 备份还原导入导出
    mongodb副本集(选举,节点设置,读写分离设置)
    mongodb副本集的内部机制(借鉴lanceyan.com)
    sqlserver 登录记录(登录触发器)
    wmic命令用法小例
    mysql查询相关的命令解析
    学习笔记:APP 瘦身 & 增加bitcode支持编译第三方框架
    关于Git的一些学习笔记
    [转]Xcode中LLDB的使用
    Swift学习笔记(2):willSet与didSet
  • 原文地址:https://www.cnblogs.com/Wurq/p/3750276.html
Copyright © 2011-2022 走看看