UVA - 146 - ID Codes （枚举排列）

UVA - 146

ID Codes

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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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 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

Source

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 2. Data Structures and Libraries :: Data Structures With Built-in Libraries ::  STL algorithm
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Data Structures and Libraries :: Linear Data Structures with Built-in Libraries ::  C++ STL algorithm (Java Collections)
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 3. Brute Force ::  Elementary Skills
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Data Structures and Libraries :: Linear Data Structures with Built-in Libraries :: C++ STL algorithm (Java Collections)

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思路：STL中next_permutation的使用，生成当前序列按照字典序排列的的下一个序列，对于字符串也可以，对于可重集也行，要注意：就这个题目而言，如果当前序列是最后一个序列的话，就输出No Successor，所以用next_permutation之前还要判断一下是否是最后一个序列。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

char str[55];

int fun(char a[]) {
	int len = strlen(a);
	for(int i = 0; i < len - 1; i++) {
		if(a[i] < a[i + 1]) return 1;
	}
	return 0;
}

int main() {
	while(scanf("%s", str), str[0] != '#') {
		int len = strlen(str);
		if(fun(str)) {
			next_permutation(str, str + len);
			printf("%s\n", str);
		}
		else printf("No Successor\n");
	}
	return 0;
} 

更新：原来next_permutation本身可以判断是否到了最后一个按照字典序排列的序列，所以直接用next_permutation即可

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

char str[55];

int main() {
	while(scanf("%s", str), str[0] != '#') {
		int len = strlen(str);
		if(next_permutation(str, str + len))
			printf("%s\n", str);
		else printf("No Successor\n");
	}
	return 0;
}