poj 1731 Orders(暴力)

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 = 200 + 10;
void PrintPermu( int n, char P[], char A[], int cur )
{
    int i, j;

    if ( cur == n )
    {
        for ( i = 0; i < n; ++i )
            cout << A[i];
        cout << endl;
    }
    else
    {
        for ( int i = 0; i < n; ++i )
        {
            if ( i == 0 || 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];
                    PrintPermu( n, P, A, cur + 1 );
                }
            }
        }
    }
}

int main()
{
    char P[MAX_N], A[MAX_N];

    cin >> P;
    sort( P, P + strlen(P) );
    PrintPermu( strlen(P), P, A, 0 );
    return 0;
}