acm pku 1056 IMMEDIATE DECODABILITY的代码实现

 

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

Source

Pacific Northwest 1998

 

       这道题目事实上，就是求编码的有效性，或者说是该编码是否可以进行译码。其算法思想如下：判断某编码群中，任意一个编码是否是其它任意编码的前缀，这个问题是容易的。举个例子，判断Code[i]是否是Code[j]的前缀，用函数strstr(Code[i], Code[j]) ?= &Code[i][0]实现。

 

具体实现如下：

#include "iostream"

using namespace std;

 

char Code[10][10];

int SetNum = 0;

 

void Judge(int CodeNum)

{

       bool bjudge = true;

       int i, j;

 

       SetNum ++;

       for(i = 0; i < CodeNum; i ++)

       {

              for(j = i+1; j < CodeNum; j ++)

              {

                     if(strstr(Code[i], Code[j]) == &Code[i][0] || strstr(Code[j], Code[i]) == &Code[j][0])

                     {

                            bjudge = false;

                            break;

                     }

              }

       }

 

       if(bjudge)

       {

              cout << "Set " << SetNum << " is immediately decodable" <<endl;

       }

       else

       {

              cout << "Set " << SetNum << " is not immediately decodable" <<endl;

       }

}

 

int main(void)

{

       int i, CodeNum;

      

       while(cin>>Code[0])

       {

              i = 0;

              CodeNum = 0;

              while(strcmp(Code[i], "9") != 0)

              {

                     CodeNum ++;

                     i ++;

                     cin >> Code[i];

              }

              Judge(CodeNum);

       }

 

       return 0;

}

 

执行结果：

Problem: 1056

 

User: uestcshe

Memory: 192K

 

Time: 0MS

Language: C++

 

Result: Accepted