搜索系列——Anagram（全排列）

Anagram
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 16 Accepted Submission(s) : 3
Problem Description
You are to write a program that has to generate all possible words from a given set of letters.
Example: Given the word "abc", your program should - by exploring all different combination of the three letters - output the words "abc", "acb", "bac", "bca", "cab" and "cba".
In the word taken from the input file, some letters may appear more than once. For a given word, your program should not produce the same word more than once, and the words should be output in alphabetically ascending order.

Input
The input consists of several words. The first line contains a number giving the number of words to follow. Each following line contains one word. A word consists of uppercase or lowercase letters from A to Z. Uppercase and lowercase letters are to be considered different. The length of each word is less than 13.

Output
For each word in the input, the output should contain all different words that can be generated with the letters of the given word. The words generated from the same input word should be output in alphabetically ascending order. An upper case letter goes before the corresponding lower case letter.

Sample Input
3
aAb
abc
acba

Sample Output
Aab
Aba
aAb
abA
bAa
baA
abc
acb
bac
bca
cab
cba
aabc
aacb
abac
abca
acab
acba
baac
baca
bcaa
caab
caba
cbaa

Source
PKU

还是一题回溯加dfs，题目要求是把字母全排列，据说C++有个函数能够实现，但这里要求按字母表排列，而且在同一个字母的条件下，大写比小写要排前。于是我就用纯C做了。
我的思路是构造两个数组ch和val来储存字母和字母的值，字母的值转化成数值储存，比如a是0.5， b是1.5，A是0，B是1，然后再从小到大排序，然后dfs前序遍历输出就是题目要求的输出了。

但是，编完后发现一个很大的问题，在有重复字母的情况下会输出重复的排列。。。
代码如下（未AC）：

[cpp]
#include<stdio.h>
#include<string.h>

float val[13];
int num, res[13];
char ch[13];

void dfs(int cur)
{
    if (cur == num)
    {
        for (int i = 0; i < num; i++)
            printf("%c", ch[res[i]]);
        printf("\n");
        return;
    }
    for (int i = 0; i < num; i++)
    {
        int ok = 1;
        res[cur] = i;
        for (int j = 0; j < cur; j++)
            if (res[j] == i)
            {
                ok = 0;
                break;
            }
        if (ok)
            dfs(cur + 1);
    }
    return;
}

int main()
{
    int n;
    num = 0;
    scanf("%d", &n);
    while (n--)
    {
        num = 0;
        scanf("%s", ch);
        for (int i = 0; ch[i] != '\0'; i++)
        {
            num++;
            if (ch[i] >= 'A'&& ch[i] <= 'Z')
                val[i] = ch[i] - 'A';
            else
                val[i] = ch[i] - 'a' + 0.5;
        }
        for (int i = 0; i < num; i++)
            for (int j = i + 1; j < num; j++)
                if (val[i] > val[j])
                {
                    float e = val[i];
                    val[i] = val[j];
                    val[j] = e;
                    char t = ch[i];
                    ch[i] = ch[j];
                    ch[j] = t;
                }
        dfs(0);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>

float val[13];
int num, res[13];
char ch[13];

void dfs(int cur)
{
if (cur == num)
{
  for (int i = 0; i < num; i++)
   printf("%c", ch[res[i]]);
  printf("\n");
  return;
}
for (int i = 0; i < num; i++)
{
  int ok = 1;
  res[cur] = i;
  for (int j = 0; j < cur; j++)
   if (res[j] == i)
   {
    ok = 0;
    break;
   }
  if (ok)
   dfs(cur + 1);
}
return;
}

int main()
{
int n;
num = 0;
scanf("%d", &n);
while (n--)
{
  num = 0;
  scanf("%s", ch);
  for (int i = 0; ch[i] != '\0'; i++)<

补充：软件开发 , C++ ,