hdu 1305(字典树)注意本题的输入！！！

Immediate Decodability

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 852    Accepted Submission(s): 453

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

0110001000009011001000009

 

Sample Output

Set 1 is immediately decodableSet 2 is not immediately decodable

//Trie树的性质：//1. 根节点不包含字符，除根节点外每一个节点都只包含一个字符。 //2. 从根节点到某一节点，路径上经过的字符连接起来，为该节点对应的字符串。 //3. 每个节点的所有子节点包含的字符都不相同。#include<stdio.h>#include<string.h>#include<stdlib.h>typedef struct Node{    struct Node *child[2];    int flag;//终止位置的标记}Node,*Trie;int path;char str[19];Trie root;Trie newnode(){//生成一个新的结点并进行初始化    Trie temp;    temp=(Trie)malloc(sizeof(Node));    for(int i=0;i<2;i++)        temp->child[i]=NULL;    temp->flag=0;    return temp;}void Insert(){    int i,len,index;    Trie curn,newn;    len=strlen(str);    curn=root;    for(i=0;i<len;i++)    {        index=str[i]-'0';        if(curn->child[index]==NULL)//该字符没有出现过        {            path=0;//此路线未曾走过            newn=newnode();            curn->child[index]=newn;        }        curn=curn->child[index];//变换当前节点        if(curn->flag)return;//边插入边判断    }    curn->flag=1;}int main(){    int result,cases;    result=1;    cases=1;    root=newnode();    while(scanf("%s",str)!=EOF)    {        if(str[0]=='9')        {            if(result)                printf("Set %d is immediately decodable\n",cases++);            else                printf("Set %d is not immediately decodable\n",cases++);            root=newnode();            result=1;            continue;        }        path=1;        Insert();        if(path==1)            result=0;    }    return 0;}