字典树——Immediate Decodability

 Immediate Decodability

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

  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

用字典树的思想构建的二叉树，在构建的过程中，判断是不是之前的前缀，输出。

#include <stdio.h>#include <string.h>#include <stdlib.h>struct Tire{    struct Tire *a[2];    int flag;    int show;}*head,*now;Tire* lianj(){    Tire *p;    p = new Tire;    p->flag = 0;    p->show = 1;    for(int i = 0; i<2; i++)        p->a[i] = NULL;    return p;}int goujian(char*st){    char *p = st;    int i = 0,cha;    while(*p!='\0')    {        if(*p == '0')            cha = 0;        else            cha = 1;        if(now->flag == 1)            return 0;        if(now->a[cha] == NULL)        {            now->a[cha] = lianj();        }        now ->show += 1;        now = now->a[cha];        p++;    }    if(now->show == 1)    {        now->flag = 1;        return 1;    }    return 0;}int main(){    head = new Tire;    now = head;    head = lianj();    int i = 1,bo = 1;    char st[1000];    while(~scanf("%s",st))    {        now = head;        if(strcmp(st,"9") != 0)        {            if(bo)                bo = goujian(st);        }        else        {            if(bo)                printf("Set %d is immediately decodable\n",i);            else                printf("Set %d is not immediately decodable\n",i);            i++;            bo = 1;            head = new Tire;            now = head;            head = lianj();        }    }    return 0;}