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)

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).

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.

01

10

0010

0000

9

01

10

010

0000

9

Set 1 is immediately decodable

Set 2 is not immediately decodable

 题目大意：
就是输多组0 1数据，任意一组数据不能成为另一种的前缀；

一组符号的编码是立即可解码的如果没有前缀的代码一个符号代码的另一个象征。我们将假定这个问题,所有的二进制代码,没有两个编码在一组编码是相同的,每个代码至少有一个,不超过十位,每组至少有两个编码和不超过8个。

例子:假设一个字母,符号{ A,B,C,D }

下面的代码是立即可解码

#include<iostream>

#include<string>

#include<cstring>

using namespace std;

int main()

{

 string  str[10];

   int i=0; int y=1,w=0;

    while(cin>>str[i])

    {

        if(str[i]=="9")

        {

            i=0;

            w++;

            if(y==1)cout<<"Set "<<w<<" is immediately decodable"<<endl;

            else cout<<"Set "<<w<<" is not immediately decodable"<<endl;

            y=1;

        }

         int len=str[i].length();      //新输入字符长度

        for(int j=0;j<i;j++)           //没输一组数据都与原来的进行比较

        {

            int t=0;

           int len1=str[j].length();     //已存入数据的字符长度

            for(int k=0;k<len&&k<len1;k++)    //保证以短的字符串长度为比较界限

            {

                if(str[i][k]==str[j][k])

                {

                        t++;

                }

            }

            if(t==min(len,len1)){y=0;break;}  //若等式成立则说明此字符串必为某字符串前缀

        }

        i++;

    }

    return 0;

}