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前缀匹配

2024-07-15 04:09:56   219人阅读

Problem B

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 14   Accepted Submission(s) : 12
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 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
 
 
借鉴别人的代码：
#include<cstdio> 
#include<cstring> 
   
char s[1000][100]; 
int n; 
   
inline bool check() 
{ 
    for (int i = 0; i < n - 1; i++) 
        for (int j = i + 1; j < n; j++) 
        { 
            bool flag = false; 
            int l1 = strlen(s[i]); 
            int l2 = strlen(s[j]); 
            for (int k = 0; k < l1 && k < l2; k++)///比strstr()要快一点 
                if (s[i][k] != s[j][k]) 
                { 
                    flag = true; 
                    break; 
                } 
            if (!flag) return false; 
        } 
    return true; 
} 
   
int main(void) 
{ 
    int cas = 0; 
    while (gets(s[0])) 
    { 
        for (n = 1;; n++) 
        { 
            gets(s[n]); 
            if (s[n][0] == ‘9‘) 
                break; 
        } 
        if (check()) printf("Set %d is immediately decodable\n", ++cas); 
        else printf("Set %d is not immediately decodable\n", ++cas); 
    } 
}
 
我的代码:

不能输入数据
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX 20

struct Trie{
 Trie *next[MAX];
 int v;
};
Trie *root;

void CreatTrie(char * str){
 int len=strlen(str),i,j,id;
 Trie *p=root,*q;
 for(i=0;i<len;i++){
        id=str[i]-‘0‘;
  if(p->next[id]==NULL){
   q=(Trie*)malloc(sizeof(Trie));
   q->v=1;
   for(j=0;j<len;j++){
    q->next[j]=NULL;    
   }
   p->next[id]=q;
   p=p->next[id];
  }
  else{
   p->next[id]->v++;
            p=p->next[id];
  }
 }
    p->v=-1;
}

int FindTrie(char *str){
 int len=strlen(str),i;
 Trie *p=root;
 for(i=0;i<len;i++){
    int id=str[i]-‘0‘;
    p=p->next[id];
    if(p==NULL)
     return 0;
    if(p->v==-1)
     return 1;
    }
 return 1;
}

int main()
{
 char a[20];
 int flag,k,i;
 for(i=0;i<MAX;i++)
  root->next[i]=NULL;
    while(gets(a)&&a[0]!=‘9‘){
  CreatTrie(a);
//  k++;
 }
 gets(a);
 if(FindTrie(a)) printf("YES\n");
 else printf("NO\n");
 return 0;
}

 


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