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  • HDU1305 Immediate Decodability(水题字典树)

    巧了,昨天刚刚写了个字典树,手到擒来,233。

    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
    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<memory.h>
    #include<algorithm>
    #include<cstring>
    #include<string>
    using namespace std;
    char s[20];
    int note[3000];//是否有延伸 
    int c[3000][3],cnt;//记录下一个点节点 
    bool flag=false;
    void _add()
    {
        int tmp=0;
        for(int i=0;i<strlen(s)-1;i++){
            if(c[tmp][s[i]-'0']==0) c[tmp][s[i]-'0']=++cnt;
            tmp=c[tmp][s[i]-'0'];
            if(note[tmp]) flag=true;
        }
        if(c[tmp][s[strlen(s)-1]-'0']!=0) flag=true;
        else {
           c[tmp][s[strlen(s)-1]-'0']=++cnt;
           note[cnt]=1;
        }
    }
    int main()
    {
        int T=0;
        while(~scanf("%s",s)){
            if(s[0]=='9'){
                if(!flag)printf("Set %d is immediately decodable
    ",++T);
                else printf("Set %d is not immediately decodable
    ",++T);
                flag=false;cnt=0;
                memset(c,0,sizeof(c));
                memset(note,0,sizeof(note));
            }
            else _add();
        }
        return 0;
    } 

     

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  • 原文地址:https://www.cnblogs.com/hua-dong/p/7614949.html
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