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 = ;
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 = ;
for (int i = ; 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 = ;
for (int j = ; 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 = ; i < n; ++i)
printf("%c", A[i]);
printf("\n");
}
else
{
for (i = ; i < n; ++i)
{
if (!i || P[i] != P[i-])
{
int c1 = , c2 = ; for (j = ; j < cur; ++j)
if (A[j] == P[i]) c1++;
for (j = ; j < n; ++j)
if (P[i] == P[j]) c2++; if (c1 < c2)
{
A[cur] = P[i];
PrintPermutation(n, P, A, cur + );
}
}
}
}
} int main()
{
int n;
char P[MAX_N]; cin >> n;
for (int i = ; i < n; ++i)
{
cin >> P;
SortAlp(P, strlen(P));
PrintPermutation(strlen(P), P, A, );
}
return ;
}