UVa 644 Immediate Decodability

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吐槽下我的渣渣英語啊，即使叫Google翻譯也沒有看懂，最後還是自己讀了好幾遍題才讀懂。

題目大意：題意很簡單，就是給一些互不相同的由‘0‘，‘1‘組成的字串，看看有沒有一個字串是否會成為另一個的開頭的子串。

直接簡單粗暴的去比較就可以了。

這是原題：

Immediate Decodability

 

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 a data file. 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.

 

The Sample Input describes the examples above.

 

Sample Input 

 

0110001000009011001000009

 

Sample Output 

Set 1 is immediately decodableSet 2 is not immediately decodable

 

 

Miguel A. Revilla 
2000-01-17

AC代碼：

 1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7  8 char code[9][11]; 9 bool cmp(char s1[], char s2[]);10 11 int main(void)12 {13     #ifdef LOCAL14         freopen("644in.txt", "r", stdin);15     #endif16     int kase = 0, n;17     while(gets(code[0]))18     {19         bool flag = false;20         int i, j;21         n = 0;22         while(code[n][0] != ‘9‘)23             gets(code[++n]);24 25         for(i = 0; i < n - 1; ++i)26         {27             if(flag)28                 break;29             for(j = i + 1; j < n; ++j)30             {31                 flag = cmp(code[i], code[j]);32                 if(flag)    break;33             }34         }35 36         if(!flag)37             printf("Set %d is immediately decodable\n", ++kase);38         else39             printf("Set %d is not immediately decodable\n", ++kase);40     }41     return 0;42 }43 //比較一個字串是否會成為另一個的開頭的子串44 bool cmp(char s1[], char s2[])45 {46     int l1 = strlen(s1);47     int l2 = strlen(s2);48     int lmin = min(l1, l2), i = 0;49     if(l1 == l2)//長度相等，必然不會是子串50         return false;51     for(i = 0; i < lmin; ++i)52     {53         if(s1[i] != s2[i])54             break;55     }56     if(i == lmin)57         return true;58     return false;59 }

代碼君