(UVA)644 (POJ)1056 (HDU)1305 - Immediate Decodability【字典树】

本文介绍了一种算法,用于判断一组二进制编码是否为立即可解码的符号集,即任一编码都不是其他编码的前缀。通过两种不同的实现方式——数组字典树和指针字典树,来解决这一问题,并给出示例输入输出。
IMMEDIATE DECODABILITY

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

我的代码(数组字典树):
查找和插入放在主函数一次性完成了。
#include <bits/stdc++.h>  
  
using namespace std;  
  
struct node  
{  
    int next[2];  
    int cnt;  
    bool isWord;  
    void init()   
    {
        cnt=0;  
        isWord=false;  
        memset(next,-1,sizeof(next));  
    }  
}T[100000];  
  
int main()  
{  
    int kase=0,tot=1,flag=1; char s[100]; T[0].init();  
    while(~scanf("%s",s))  
    {  
        int len=strlen(s);  
        if(s[0]!='9')
        {  
            int i,p=0;  
            for(i=0;i<len;i++)
            {  
                int x=s[i]-'0';  
                if(T[p].next[x]==-1)
                {  
                    T[tot].init();  
                    T[p].next[x]=tot++;  
                }  
                p=T[p].next[x];  
                T[p].cnt++;  
                if(T[p].isWord) flag=0;       //包含其它串  
            }  
            T[p].isWord=true;  
            if(T[p].cnt>1) flag=0;            //包含于其它串  
        }else{  
            printf("Set %d is%simmediately decodable\n",++kase,flag?" ":" not ");  
            T[0].init(); tot=1,flag=1;  
        }  
    }  
    return 0;  
}  








指针字典树:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
using namespace std;

const int MAXN = 2;
typedef struct Trie_Node
{
    bool isCode;
    struct Trie_Node *next[MAXN];
}Trie;
bool flag;

void Trie_insert(Trie *root,char *str)
{
    Trie *p=root;
    int len=strlen(str);
    for(int i=0;i<len;i++)
    {
        int k=str[i]-'0';
        if(p->next[k]==NULL)
        {
            Trie *t=new Trie;
            for(int j=0;j<MAXN;j++)
                t->next[j]=NULL;
            t->isCode=false;
            p->next[k]=t;
        }
        else
        {
            if(p->next[k]->isCode)
            {
                flag=false;
                return;
            }
        }
        p=p->next[k];
    }
    p->isCode=true;
}
void Trie_del(Trie *root)
{
    for(int i=0;i<MAXN;i++)
    {
        if(root->next[i]!=NULL)
            Trie_del(root->next[i]);
    }
    free(root);
}
char str[15];

int main()
{
    int iCase=0;
    while(scanf("%s",str)!=EOF)
    {
        iCase++;
        Trie *root=new Trie;
        for(int i=0;i<MAXN;i++)
            root->next[i]=NULL;
        root->isCode=false;
        flag=true;
        Trie_insert(root,str);
        while(scanf("%s",str))
        {
            if(str[0]=='9') break;
            if(flag) Trie_insert(root,str);
        }
        if(flag) printf("Set %d is immediately decodable\n",iCase);
        else printf("Set %d is not immediately decodable\n",iCase);
        Trie_del(root);
    }
    return 0;
}
题目数据范围小胡搞:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstdlib>

using namespace std;

int main()
{
    char str[20][20],temp[20];
    int ansNum=1,cnt=0,i,j,flag=1,len,t;
    while(~scanf("%s",temp))
    {
        if(temp[0]=='9')
        {
            for(i=1;i<cnt;i++)
            {
                if(!flag) break;
                for(j=i+1;j<=cnt;j++)
                {
                    if(!flag) break;
                    len=min(strlen(str[i]),strlen(str[j]));
                    for(t=0;t<len;t++)
                    {
                        if(str[i][t]!=str[j][t])
                        {
                            flag=1;
                            break;
                        }
                        flag=0;
                    }
                }
            }
            if(flag) printf("Set %d is immediately decodable\n",ansNum);
            else printf("Set %d is not immediately decodable\n",ansNum);

            cnt=0,flag=1,ansNum++;
            memset(str,0,sizeof(str));
        }
        else
        {
            strcpy(str[++cnt],temp);
            getchar();
        }
    }
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值