poj 1056 IMMEDIATE DECODABILITY

IMMEDIATE DECODABILITY

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9229		Accepted: 4375

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 standard 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

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>

using namespace std;

const int sonnum = 2, base = '0';
struct Trie 
{
  int num; bool terminal;
 Trie *son[sonnum];
};
Trie *NewTrie()
{
  Trie *temp = new Trie;
  temp->num = 1; temp->terminal = false;
  for (int i = 0; i < sonnum; ++i) temp->son[i] = NULL;
  return temp; 
}
bool Insert(Trie *pnt, char *s, int len)
{
  Trie *temp = pnt;
  bool mrk = true;
  for (int i = 0; i < len; ++i)
  {
    if (temp->son[s[i]-base] == NULL) temp->son[s[i]-base] = NewTrie();
    else 
    {
      temp->son[s[i]-base]->num++;
      if (temp->son[s[i]-base]->terminal == true)
        mrk = false;
    }
    temp = temp->son[s[i]-base];
  }
  temp->terminal = true;
  return mrk;
}

int main(void)
{
  Trie *tree;
  char a[20]; int cnt = 1;
#ifndef ONLINE_JUDGE
  freopen("poj2056.in", "r", stdin);
#endif
  while (~scanf("%s", a))
  {
    tree = NewTrie();
    bool flag = true;
    Insert(tree, a, strlen(a));
  while (~scanf("%s", a))
    {
      if (a[0] == '9') break;
      if (false == Insert(tree, a, strlen(a)))
        flag = false;
    }
//    if (a[0] == '9') continue;
    if (flag == false) printf("Set %d is not immediately decodable\n", cnt);
    else printf("Set %d is immediately decodable\n", cnt);
    cnt++;
}

  return 0;
}

又练习了一遍字典树，标记即可没有什么技术含量，代码还可以优化，习惯这么写了……虽然有点儿麻烦，至少是对的。