暴力--下一个排列--ID Codes

最新推荐文章于 2023-02-25 22:41:55 发布

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本文介绍了一种算法，用于生成基于特定字符集的唯一ID码的下一个排列。这些ID码由不超过50个重复的小写字母组成，并通过手术植入公民手腕上的微型计算机中。文章提供了使用C++实现的示例代码，展示了如何使用next_permutation函数来找到给定代码的下一个有效排列。

ID Codes

It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.)

An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set.

For example, suppose it is decided that a code will contain exactly 3 occurrences of `a', 2 of `b' and 1 of `c', then three of the allowable 60 codes under these conditions are:

 
        abaabc
       abaacb

 
      ababac

These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order.

Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message `No Successor' if the given code is the last in the sequence for that set of characters.

Input and Output

Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #.

Output will consist of one line for each code read containing the successor code or the words `No Successor'.

Sample input

 
  abaacb
 
 
  cbbaa
 
 
 
  #

Sample output

 
  ababac
 
 
  No Successor

题意是要求下一个排列

比如说将数字全排列

1 2 3 4 5 6

1 2 3 4 6 5

1 2 3 5 4 6

1 2 3 5 6 4

1 2 3 6 4 5

1 2 3 6 5 4.....

字母同理:

abaacb 0 1 0 0 2 1

ababac 0 1 0 1 0 2

ababca 0 1 0 1 2 0

abacab 0 1 0 2 0 1

abacba......(可转化成数字看) 0 1 0 2 1 0

那么如何求下一个排列呢?C++的STL中提供了一个库函数next_permutation,用来求下一个排列;

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,c[55],i,len;
    char s[55];
    while(gets(s)!=NULL&&s[0]!='#')
    {

        len=strlen(s);
        for(i=0; i<len; i++)
        {
            c[i]=s[i]-'a';
        }
        sort(c,c+len);
        if(next_permutation(s,s+len))
        {
            for(int i=0; i<len; i++)
                printf("%c",s[i]);
            printf("\n");
        }
        else
        printf("No Successor\n");
    }
    return 0;
}