hdu 1305 Immediate Decodability

Immediate Decodability

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5334 Accepted Submission(s): 2703

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)

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

0010

0000

010

0000

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();
    }
}