zoukankan      html  css  js  c++  java
  • POJ 1056 IMMEDIATE DECODABILITY (字典树)

    IMMEDIATE DECODABILITY
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 12972   Accepted: 6222

    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

    题意:和HDU 1671很像,就是问你有没有一个串是另一个的前缀。
    #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
    ",iCase);
            else printf("Set %d is not immediately decodable
    ",iCase);
            Trie_del(root);
        }
        return 0;
    }
     
  • 相关阅读:
    Oracle 11g服务详细介绍及哪些服务是必须开启的?
    分析函数Ratio_to_report使用
    在datagrid中,IE浏览器报错:SCRIPT5007: 无法获取属性“rowspan”的值: 对象为 null 或未定义
    身为一个产品经理应该了解自己的本职
    创意思维常用练习方法资料
    思维导图五个关键秘诀
    思维导图与超级记忆力之间的联系
    易让大脑变迟钝的九中习惯
    分享怎样快速阅读的5大方法
    思维导图运用的四个特征
  • 原文地址:https://www.cnblogs.com/wangdongkai/p/5738423.html
Copyright © 2011-2022 走看看