Immediate Decodability
Problem 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)
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
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
题目大意:
给你几段只包含0,1的序列,判断这几段序列中,是否存在至少一段序列是另一段序列的前缀。
解题分析:
只需要在每次插入字符串,并且在Trie树上创建节点的时候,判断路径上是否已经有完整的单词出现即可。
#include<iostream> #include<string.h> #include<string> char s[4005]; int tree[4005][3],vis[4005]; int len,id,root=0,t=0,num=0,flag=1; void init() { flag=1; memset(tree,0,sizeof(tree)); memset(vis,0,sizeof(vis)); } void insert() { len=strlen(s); root=0; for(int i=0;i<len;i++) { id=s[i]-'a'; if(!tree[root][id]) tree[root][id]=++num; root=tree[root][id]; if(vis[root]) flag=0; } vis[root]=1; } int main() { while(~scanf("%s",s)) { if(s[0]=='9') { if(flag==0) printf("Set %d is not immediately decodable ",++t); else printf("Set %d is immediately decodable ",++t); init(); } else insert(); } }