poj 1256 Anagram(dfs)

题目链接：http://poj.org/problem?id=1256

思路分析：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;
}