acm 试题 字符串排序

Given a string of letters(A-Z),your task is to arrange them in alphabetic order.
Following is an example:
A string "BAC" contains 3 letters B,A and C,you should output
ABC
ACB
BAC
BCA
CAB
CBA
In the output file.
A string may contain several letters same,for example "BBC" you should output like this:
BBC
BCB
CBB

Input

The first line of input contains a single integer t, the number of test cases,followed by the input data for each test data.Each test case is a string of n(1<=n<=26) letters.

Output

You should output Case K: in the first line and the sequences of arrangement in the following lines of each case.

Sample Input

2
BAC
BBC

Sample Output

Case 1:
ABC
ACB
BAC
BCA
CAB
CBA
Case 2:
BBC
BCB
CBB

 

这道题目，我采用的是递归算法实现的。 采用的链串 结构；
算法简单描述为：对一个链串，有个标记开始置换的指针，然后向后移动，直到末尾，然后置换的指针前移，重复上面的过程。如果标记的指针的内容和移动指针内容不同就打印出来；

源代码如下：

#include <stdio.h>
#include <iostream>
using namespace std ;

typedef struct str
 {
    char ch ;
    struct str * next ;
}str ;

void print(str * s); //打印
void create(str *&s);//建立
void travel(str *&s,str * r) ;//遍历，采用递归算法，实现排列问题
void swap(char &a , char & b) ;//交换

int main(int argc, char* argv[])
 {
    str * s ;
    create(s);
   
    print(s);
    travel(s,s->next);
   
    system("PAUSE");
    return 0;
}

void create(str *&s)//建立链串
 {
    s = new str ;
   
    str * p ;
    str * q = s ;
    char ch ;
   
    while((ch = getchar())!= '/n')
     {
        p = new str ;
        p->ch  = ch ;
        p->next = NULL ;
       
        q->next = p ;
        q = p ;
    }
   
    q->next = NULL ;
}

void print(str * s)
 {
    str * p ;
    p = s->next ;
    while(p != NULL)
     {
        cout<<p->ch ;
        p = p->next ;
    }
   
    putchar('/n');
}

void swap(char &a , char & b)
 {
    char temp ;
    temp = a ;
    a = b ;
    b = temp ;
}

 

void travel(str *&s,str * r) //遍历
 {
   
    str * p = r;
    str * q = p->next ;

    if(r->next == NULL)
     {
        return ;
    }

    else
     {
        while(p!= NULL)
         {
            q = p->next ;
            while(q != NULL)
             {
                swap(p->ch,q->ch);
                if(p->ch != q->ch)
                 {
                    print(s);
                    travel(s,p->next);
                }
               
                swap(p->ch,q->ch);
                q= q->next ;
            }
            p = p->next ;
           
        }
    }
}