POJ1056

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

0110001000009011001000009

Sample Output

Set 1 is immediately decodableSet 2 is not immediately decodable

题意:给你一些字符串,每组数据以9结尾,判断给出的字符串中,是否存在一个字符串是另外一个字符串的前缀(如:01为010的前缀);如果存在至少一组这种情况输出:Set %d is not immediately decodable ca;不存在则输出Set %d is immediately decodable ca。

这个题可以开两个图(map)来判断,一个map用来记字符串,另外一个图记前缀,然后判断一下就可以了(每次重新开始下一组数据的时候记得将map清空)

716K 0MS

代码如下：

#include<iostream>#include<string.h>#include<string>#include<stdio.h>#include<map>using namespace std;char a[110];int main(){    int i,j,k,ca=0;    while(1)    {        ca++;        j=1;        map<string,int> p;        map<string,int> q;        while(1)        {            if(cin>>a==0)            return 0; //一开始我傻逼的用了scanf("%s",a);结果Outpus T了            if(strcmp(a,"9")==0)                break;                if(p[a]!=0)                    j=0;                    q[a]++;                k=strlen(a);                for(i=k-1;i>0;i--)                {                    a[i]='\0';                    if(q[a]!=0)                        j=0;                    p[a]++;                }        }        if(j==1)            printf("Set %d is immediately decodable\n",ca);        else            printf("Set %d is not immediately decodable\n",ca);    }    return 0;}