zoukankan      html  css  js  c++  java
  • POJ 1056 IMMEDIATE DECODABILITY

    IMMEDIATE DECODABILITY
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 10504   Accepted: 4971

    Description

    An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight. 

    Examples: Assume an alphabet that has symbols {A, B, C, D} 

    The following code is immediately decodable: 
    A:01 B:10 C:0010 D:0000 

    but this one is not: 
    A:01 B:10 C:010 D:0000 (Note that A is a prefix of C) 

    Input

    Write a program that accepts as input a series of groups of records from standard input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

    Output

    For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

    Sample Input

    01
    10
    0010
    0000
    9
    01
    10
    010
    0000
    9
    

    Sample Output

    Set 1 is immediately decodable
    Set 2 is not immediately decodable
     
    

    Source

    字符串匹配的简单题,和UVa 644(http://www.cnblogs.com/lzj-0218/p/3526656.html)完全相同

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 
     6 using namespace std;
     7 
     8 int main()
     9 {
    10     int k=0,t=0;
    11     char s[10][20];
    12     char input[20];
    13     bool isImm=true;
    14 
    15     while(gets(input))
    16     {
    17         if(input[0]=='9')
    18         {
    19             k++;
    20             if(isImm)
    21                 printf("Set %d is immediately decodable
    ",k);
    22             else
    23                 printf("Set %d is not immediately decodable
    ",k);
    24             isImm=true;
    25             t=0;
    26         }
    27         else if(isImm)
    28         {
    29             bool diff=true;
    30             for(int i=0;diff&&i<t;i++)
    31             {
    32                 int j;
    33                 int len=min(strlen(s[i]),strlen(input));
    34                 for(j=0;j<len;j++)
    35                 {
    36                     if(s[i][j]!=input[j])
    37                         break;
    38                 }
    39                 if(j==len)
    40                     diff=isImm=false;
    41             }
    42             if(diff)
    43             {
    44                 strcpy(s[t],input);
    45                 t++;
    46             }
    47         }
    48 
    49 
    50     }
    51 
    52     return 0;
    53 }
    [C++]
  • 相关阅读:
    kafka参数解析+启动参数解析
    flume参数解析+启动参数解析
    flume+flume+kafka消息传递+storm消费
    C++快速读取大文件
    阿里云ecs : Couldn't connect to host, port: smtp.aliyun.com, 25; timeout -1;
    对渲染相关操作封装的类库
    51nod-1134 最长递增子序列,用线段树将N^2的dp降到NlogN
    51nod-1503 猪和回文
    51nod1073-约瑟夫环,递归。
    算法基础题
  • 原文地址:https://www.cnblogs.com/lzj-0218/p/3537109.html
Copyright © 2011-2022 走看看