POJ 1146 ID Codes (生成排列)

ID Codes

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6199 Accepted: 3712

Description

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

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

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

Sample Input

abaacbcbbaa#

Sample Output

ababacNo Successor

Source

New Zealand 1991 Division I,UVA 146

题目链接：http://poj.org/problem?id=1146

题目大意：按字典序升序生成下一个排列，若当前已经是字典序最大的排列则输出No Successor

题目分析：按字典序升序生成下一个排列的步骤：
1) 找最后一个正序：t1 = max { ( j | sj-1 >= sj ) }
2) t1 = -1则说明当前为最大，退出输出No Successor
3) 否则，找最后大于st1-1者st2：t2 = { k | st1-1 < sk}
4) 互换st1-1和st2
5) 反排st2后面的数

#include <cstdio>  #include <cstring>  #include <algorithm>  using namespace std;  char s[55];    bool get()  {      int l = strlen(s), t1 = -1;      for(int i = l - 1; i > 0; i--)          if(s[i - 1] < s[i])           {              t1 = i;  //找到最后一个升序t1              break;           }      if(t1 == -1)          return false;      int t2 = t1;      for(int i = t1 + 1; i < l; i++)      {          if(s[t1 - 1] >= s[i])              continue;           t2 = i;  //找到t1后最后一个大于st1-1的数t2     }      swap(s[t1 - 1], s[t2]); //交换st1-1和st2      sort(s + t1, s + l); //反排交换后st2后的数      return true;  }    int main()  {      while(scanf("%s", s) != EOF && s[0] != '#')          printf("%s\n", get() ? s : "No Successor");  }