UVA 644 - Immediate Decodability（字符串）

Immediate Decodability 

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 a data file. 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.

The Sample Input describes the examples above.

Sample Input 

0110001000009011001000009

Sample Output 

Set 1 is immediately decodableSet 2 is not immediately decodable

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Miguel A. Revilla 
2000-01-17

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如果出现某个字符串与其他的字符串的前几位完全相同，则不可立即解码，否则可以

一直输入，遇到9就处理。把所有的字符串用qsort排个序，不可立即解码的两个字符串一定挨着所以i和i+1相比较即可

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char str[111][111];int cmp(const void *a,const void *b){    return strcmp((char*)a,(char*)b);}int main(){    int t=1;    int k=0,flag;    while(gets(str[k++])!=NULL)    {        if(str[k-1][0]=='9')        {            qsort(str,k,sizeof(str[0]),cmp);            for(int i=0; i<k-1; i++)            {                flag=1;                for(int j=0; j<strlen(str[i])&&j<strlen(str[i+1]); j++)                {                    if(str[i][j]!=str[i+1][j])                    {                        flag=0;                        break;                    }                }                if(flag)                {                    printf("Set %d is not immediately decodable\n",t++);                    k=0;                    break;                }            }            if(!flag) printf("Set %d is immediately decodable\n",t++);            k=0;        }    }    return 0;}