杭电acm 1305Immediate Decodability（字典树）

Immediate Decodability

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3590    Accepted Submission(s): 1865

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

  

   01
10
0010
0000
9
01
10
010
0000
9
  

 

Sample Output

  

   Set 1 is immediately decodable
Set 2 is not immediately decodable
  

 

Source

Pacific Northwest 1998

想法：

字典树模板

代码：

#include<stdio.h>
#include<string.h>
const int N=2;
struct Trie
{
    int num;
    Trie *nx[N];
};
Trie *root;
void init (Trie *t)
{
    for (int j = 0; j < N; ++j) //初始化
    {
        t->nx[j] = NULL;
    }
    t->num = 0;
}
void Insert (char str[])//插入单词
{
    Trie *p = root;
    int len = strlen (str);
    for (int i = 0; i < len; ++i)
    {
        int id = str[i] - '0';
        if (p->nx[id] == NULL)
        {
            Trie *t = new Trie;
            init(t);
            p->nx[id] = t;
        }
        p = p->nx[id];
        p->num++;
    }
}
int Find(char str[])//查找前缀为str的单词数
{
    Trie *p = root;
    int cnt,len = strlen(str);
    for (int i = 0;i < len;++i)
    {
        int id = str[i] - '0';
        if (p->nx[id] == NULL) return 0;
        p = p->nx[id];
        cnt = p->num;
    }
    return cnt;
}
void Delete(Trie *t)
{
    for (int i = 0;i < N;++i)
    {
        if (t->nx[i]!=NULL) Delete(t->nx[i]);
    }
    delete t;
}
int main()
{
    int kas=0;
    root = new Trie;
    init(root);
    int kk=1;
    char s[50][50];
    while(gets(s[++kas]))
    {
        if(s[kas][0]=='9')
        {
           int i,flag=1;
           for(i=1;i<kas;i++)
           {
               if(Find(s[i])>=2)
               {
                   flag=0;
                   break;
               }
           }
           if(flag) printf("Set %d is immediately decodable\n",kk++);
           else  printf("Set %d is not immediately decodable\n",kk++);
           kas=0;
            root = new Trie;
            init(root);
        }
        else
        {
            Insert(s[kas]);
        }
    }
}