poj 1146 ID Codes (字符串处理 生成排列组合 生成当前串的下一个字典序排列 【*模板】 )

ID Codes

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6229		Accepted: 3737

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

abaacb
cbbaa
#

Sample Output

ababac
No Successor

Source

New Zealand 1991 Division I,UVA 146

 

题目分析：给你一个字符串，长度不超过50. 现在要让你输出该串的下一个字典序排列。当然前提是存在就输出，否则就输出“No Successor”。

参考书籍《算法设计编程实验》(P 120)

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;
char s[100];
int len;

bool get()
{
	len=strlen(s);
	int i=len-1;
	while(i>0 && s[i-1]>=s[i] )
		i--; //找到最后一个正序
	if(i==0)
		return false; //表示当前排列为最后一个排列，则返回false
	int pos=i;
	for(int j=i+1; j<len; j++)
	{
		if(s[j]<=s[i-1] )
			continue;
		if(s[j] < s[pos] )
			pos=j;
	}
	swap(s[pos], s[i-1]);
	sort(s+i, s+len);
	return true;
}

int main()
{
	//判断一个序列还有没有下一个全排列序列 如果存在就输出

	while(scanf("%s", s)!=EOF )
	{
		if(s[0]=='#') break;
		if( get()==true )
			printf("%s\n", s);
		else
			printf("No Successor\n");

	}
	return 0;
}

分析：利用该算法，也可以一直输出下去，从串的当前字典位置开始，直到输出至最后一个字典序。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;
char s[100];
int len;

bool get()
{
    len=strlen(s);
    int i=len-1;
    while(i>0 && s[i-1]>=s[i] )
        i--; //找到最后一个正序
    if(i==0)
        return false; //表示当前排列为最后一个排列，则返回false
    int pos=i;
    for(int j=i+1; j<len; j++)
    {
        if(s[j]<=s[i-1] )
            continue;
        if(s[j] < s[pos] )
            pos=j;
    }
    swap(s[pos], s[i-1]);
    sort(s+i, s+len);
    return true;
}

int main()
{
    //判断一个序列还有没有下一个全排列序列 如果存在就输出

    while(scanf("%s", s)!=EOF )
    {
        if(s[0]=='#') break;
        if(get()==true){
                printf("%s\n", s);
            while( get()==true )
                printf("%s\n", s);
        }
        else
            printf("No Successor\n");
    }
    return 0;
}

还可以这样变体：将初始输入的字符串进行排序：然后在利用上面的代码，输出当前串的所有排列

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;
char s[100];
int len;

bool get()
{
    int i=len-1;
    while(i>0 && s[i-1]>=s[i] )
        i--; //找到最后一个正序
    if(i==0)
        return false; //表示当前排列为最后一个排列，则返回false
    int pos=i;
    for(int j=i+1; j<len; j++)
    {
        if(s[j]<=s[i-1] )
            continue;
        if(s[j] < s[pos] )
            pos=j;
    }
    swap(s[pos], s[i-1]);
    sort(s+i, s+len);
    return true;
}

int main()
{
    //判断一个序列还有没有下一个全排列序列 如果存在就输出

    while(scanf("%s", s)!=EOF )
    {
        if(s[0]=='#') break;
        len=strlen(s);
        sort(s, s+len);
        printf("%s\n", s);

        if(get()==true){
                printf("%s\n", s);
            while( get()==true )
                printf("%s\n", s);
        }
        else
            printf("No Successor\n");
    }
    return 0;
}