HDOJ Immediate Decodability

 

Immediate Decodability

http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=12654&pid=1003

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 2

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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

 下面的代码是错误的，如果输入 

010

01

这组数据的输入结果显然不对，幸运的是，这个错误现在找到了。

水题，直接用字典树的模板就可以做出，但不要局限于某种结构，下面是我用两种方法实现的，写代码的时候要特别注意细节。

用了85行，怪可怜的额。

#include<iostream>
#include<cstring>
using namespace std;
struct btree{
       struct btree *child[2]; 
       int n; //用来判断这个点是不是被某个串经过,经过为1
};
char save[10][12];  //保存
struct btree *root;
void insert(char *source){
    int len,i,j;
    len=strlen(source);
    if(len==0)  return ;
    struct btree *current,*newnode;
    current=root;
    for(i=0;i<len;i++){
         if(current->child[source[i]-'0']!=NULL){
              current=current->child[source[i]-'0'];
                 current->n=1; //把经过的点置为1
         }
         else{
             newnode=new btree;
             newnode->child[0]=newnode->child[1]=NULL;
             current->child[source[i]-'0']=newnode;
             current=newnode; 
             current->n=1;                
         }                                        
    }
    current->n=0;  //避免自身的影响所以必须置0
}
int find(char *source){
    int i,len;
    len=strlen(source);
    if(len==0)  return 0;
    struct btree *current;
    current=root;
    for(i=0;i<len;i++){
        if(current->child[source[i]-'0']!=NULL)
            current=current->child[source[i]-'0'];          
        else
           return 0;                  
    }
    if(current->n==1) //该串的最后一个字符在另一个串上，则说明该串是另一个串的子串
         return 0;
    return 1;
}
void destroy(struct btree *p){ //把以前建立的节点抹掉
     struct btree *q,*h;
     if(p!=NULL){  //这个if不能省略 
       q=p->child[0];
       h=p->child[1];
       p=NULL; //这里怎么我用free(p); 和 delete  p ; 就不可以呢？纠结啊
       destroy(q);
       destroy(h);
     }
}
int main(){
    char temp[11];
    int cnt=0,j=0,flag;
    root=new btree;
    root->n=0;
    root->child[0]=root->child[1]=NULL;
    while(scanf("%s",temp)!=EOF){
          if(strcmp(temp,"9")==0){
              flag=1;
              for(int i=0;i<j;i++){                
                      if(find(save[i])==0)
                         flag=0;
              }                    
               if(flag==1)
                  printf("Set %d is immediately decodable\n",++cnt);
               else{
                  printf("Set %d is not immediately decodable\n",++cnt);
                  flag=1;  //注意要恢复为1
               }
               j=0; 
               destroy(root->child[0]); //释放左子树空间
               destroy(root->child[1]);   //释放右子树空间      
               continue;                  
          }
          strcpy(save[j++],temp);
          insert(temp);                                                                                                               
    }
    return 0;
}

 下面这个只用了38行，效率还好点。     

//两两验证一个串是不是另一个串的前缀，注意是前缀不是子串
#include<iostream>
#include<cstring>
using namespace std;
int main(){
    char temp[15], s[15][15],a[15];
    int cnt=0,flag,h=0;
    while(scanf("%s",temp)!=EOF){
         if(strcmp(temp,"9")==0){
             flag=1;
             for(int i=0;i<h-1;i++){    //两两比较             
                 for(int j=i+1;j<h;j++){
                     if(strlen(s[j])<strlen(s[i]) ){ 
            //从长度长的串中拷贝出长度为短的字符串到a                          
                         strncpy(a,s[i],strlen(s[j]));
                         a[strlen(s[j])]='\0';
                         if(strcmp(a,s[j])==0)
                            {  flag=0;  break; }
                     }
                     else{
                         strncpy(a,s[j],strlen(s[i]));
                         a[strlen(s[i])]='\0';
                         if(strcmp(a,s[i])==0)
                             {  flag=0; break;  }
                     }                   
                 }
             }                        
              if(flag==1)
                  printf("Set %d is immediately decodable\n",++cnt);
              else
                  printf("Set %d is not immediately decodable\n",++cnt);                                 
               h=0;  //注意要恢复为0
               memset(s,0,sizeof(s));
               continue;                         
         }                                                      
         strcpy(s[h++],temp);                                                       
    }
    return 0;
}