POJ 1256：Anagram

Anagram
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 18393 Accepted: 7484

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

Hint
An upper case letter goes before the corresponding lower case letter.
So the right order of letters is ‘A’<’a’<’B’<’b’<…<’Z’<’z’.

题意就是按照字典顺序输出所有字符串。
正常来说，只需要一个sort，一个next_permutation就OK的题，唯一一个值得说明的就是要把A，B，C这样的大写字母穿插到里面，要不然出来的顺序不会是AaBbCc，而是ABCabc。所以需要自己写一个自定义函数cmp用来比较。我这里为了穿插进去，使用了ASCII码+31.5，就使得每一个大写字母刚好穿进了小写字母中。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
using namespace std;

char s[5000];

bool cmp(char a,char b)
{
    double front,behind;

    if(a>='A'&&a<='Z')
        front = (double)a+31.5;
    else
        front = (double)a;

    if(b>='A'&& b<='Z')
        behind = (double)b+31.5;
    else
        behind = (double)b;

    return front<behind;
}
int main()
{
    int count;
    cin>>count;

    while(count--)
    {
        cin>>s;
        sort(s,s+strlen(s),cmp);

        do
        {
            cout<<s<<endl;
        }while(next_permutation(s,s+strlen(s),cmp));
    }

    return 0;
}

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