poj 1256 Anagram

题目链接：http://poj.org/problem?id=1256

思路：

    该题为含有重复元素的全排列问题；由于题目中字符长度较小，采用暴力法解决。

代码如下：

#include <iostream>
#include <algorithm>
using namespace std;

const int MAX_N = 20;
char P[MAX_N], A[MAX_N];

char * SortAlp( char P[], int n )
{
    int Low[MAX_N], Upper[MAX_N];
    int LowLen, UpperLen;

    LowLen = UpperLen = 0;
    for ( int i = 0; i < n; ++i )
    {
        if ( ‘A‘ <= P[i] && P[i] <= ‘Z‘ )
            Upper[UpperLen++] = P[i];
        else
            Low[LowLen++] = P[i];
    }

    sort( Low, Low + LowLen );
    sort( Upper, Upper + UpperLen );

    int Index_L, Index_U;
    Index_L = Index_U = 0;
    for ( int j = 0; j < n; ++j )
    {
        if (Upper[Index_U] - ‘A‘ + ‘a‘ <= Low[Index_L]
            && Index_U < UpperLen)
            P[j] = Upper[Index_U++];
        else
            P[j] = Low[Index_L++];
    }

    return P;
}

void PrintPermutation( int n, char P[], char A[], int cur )
{
    int i, j;

    if ( cur == n )
    {
        for ( i = 0; i < n; ++i )
            printf( "%c", A[i] );
        printf("\n");
    }
    else
    {
        for ( i = 0; i < n; ++i )
        {
            if ( !i || P[i] != P[i-1] )
            {
                int c1 = 0, c2 = 0;

                for ( j = 0; j < cur; ++j )
                    if ( A[j] == P[i] ) c1++;
                for ( j = 0; j < n; ++j )
                    if ( P[i] == P[j] ) c2++;

                if ( c1 < c2 )
                {
                    A[cur] = P[i];
                    PrintPermutation( n, P, A, cur + 1 );
                }
            }
        }
    }
}

int main()
{
    int n;
    char P[MAX_N];

    cin >> n;

    for ( int i = 0; i < n; ++i )
    {
        cin >> P;
        SortAlp( P, strlen(P) );
        PrintPermutation( strlen(P), P, A, 0 );
    }

    return 0;
}