Immediate Decodability

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3444 Accepted Submission(s): 1791

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

Sample Output

Set 1 is immediately decodable

Set 2 is not immediately decodable

題目大意：給你一些二進制組成的密碼，詢問是否可以解碼，規定如果這些密碼裏存在某些二進制密碼是另外一些二進制密碼的前綴，則不可以解碼，如果不是則代表可以解碼，以數字9代表當前一組數據輸入結束。

解題思路：沒得到一組密碼，就將該密碼插入到字典樹裏面去，因爲這裏只有0和1，所以密碼樹只有兩個子樹1和0，當存在情況爲當前密碼存儲的位置被標記過時，則說明該位置存在密碼，即存在前綴情況，結束。

ac代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
using namespace std;
const int M=100005;
struct node
{
    int cnt;
    node *child[2];
    node()
    {
        cnt=0;
        for(int i=0; i<2; i++)
            child[i]=NULL;
    }
};
char a[15];
int flag;
node *root,*cur,*newnode;
void insert(char *a)
{
    cur=root;
    int len=strlen(a);
    for(int i=0; i<len ; i++)
    {
        int index=a[i]-'0';
        if(cur->child[index]!=NULL)
        {
            cur=cur->child[index];
            if(cur->cnt==1 || i==len-1)
            {
                flag=0;//標記 
                break;
            }
        }
        else
        {
            newnode=new node;
            cur->child[index]=newnode;
            cur=newnode;
        }
    }
    cur->cnt=1;
}
void del(node *head)
{
    for(int i=0; i<2; i++)
        if(head->child[i]!=NULL)
            del(head->child[i]);
    delete(head);
}
int main()
{
    int g=1;
    while(scanf("%s",a)!=EOF)
    {
        flag=1;//標記默認爲1 
        root=new node;
        insert(a);
        while(scanf("%s",a)!=EOF)
        {
            if(strcmp(a,"9")==0)break;
            {
                if(!flag)
                    continue;
                insert(a);
            }
        }
        if(flag)printf("Set %d is immediately decodable\n",g++);
        else
            printf("Set %d is not immediately decodable\n",g++);
        del(root);
    }
    return 0;
}

題目鏈接：點擊打開鏈接http://acm.hdu.edu.cn/showproblem.php?pid=1305

有關字典樹的有關知識請看我的字典樹專題博客：點擊打開鏈接http://blog.csdn.net/wang_heng199/article/details/76448804