题目链接：http://poj.org/problem?id=1731
思路分析：含有重复元素的全排列问题；元素个数为200个，采用暴力枚举法。
代码如下：
#include <iostream>
#include <algorithm>
using namespace std; const int MAX_N = + ;
void PrintPermu( int n, char P[], char A[], int cur )
{
int i, j; if ( cur == n )
{
for ( i = ; i < n; ++i )
cout << A[i];
cout << endl;
}
else
{
for ( int 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];
PrintPermu( n, P, A, cur + );
}
}
}
}
} int main()
{
char P[MAX_N], A[MAX_N]; cin >> P;
sort( P, P + strlen(P) );
PrintPermu( strlen(P), P, A, );
return ;
}