算法分析之有重复元素的排列问题O（n！）

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

template<class eT>
void quicksort(vector<eT>& a, int left, int right, vector<int>& next)
{
	int L = left, R = right;
	eT T = a[L];
	while (L < R) {
		while (a[R] >= T && L < R) --R;
		a[L] = a[R];
		while (a[L] < T && L < R) ++L;
		a[R] = a[L];
	}

	int M1 = L;
	R = right;
	while (L < R) {
		while (a[R] > T && L < R) --R;
		a[L] = a[R];
		while (a[L] <= T && L < R) ++L;
		a[R] = a[L];
	}
	a[L] = T;
	int M2 = L;

	for (int i = M1; i <= M2; ++i)
		next[i] = M2 + 1;

	if (left < M1 - 1)
		quicksort(a, left, M1 - 1, next);
	else if (left == M1 - 1)
		next[left] = M1;
	if (M2 + 1 < right)
		quicksort(a, M2 + 1, right, next);
	else if (M2 + 1 == right)
		next[right] = right + 1;
}

int arrange(vector<char>& elems, vector<int>& next, vector<int>& used, vector<char>& rec)
{
	int i = 0, n = elems.size();
	while (i < n && used[i]) ++i;
	if (i >= n) {
		for (i = 0; i < n; ++i)
			cout << rec[i];
		cout << endl;
		return 1;
	}

	int cnt = 0;
	while (i < n) {
		rec.push_back(elems[i]);
		used[i] = 1;
		cnt += arrange(elems, next, used, rec);
		rec.erase(rec.end()-1);
		used[i] = 0;
		
		i = next[i];
		while (i < n && used[i]) ++i;
	}
	return cnt;
}

int main(void)
{
	int n;
	while (cin >> n) {
		while (getchar() != '\n');
		vector<char> elems;
		for (int i = 0; i < n; ++i) {
			elems.push_back(getchar());
		}
		while (getchar() != '\n');

		vector<int> next(n, -1);
		quicksort(elems, 0, n - 1, next);

		// display
		for (int i = 0; i < n; ++i)
			cout << elems[i] << " ";
		cout << endl;
		for (int i = 0; i < n; ++i)
			cout << next[i] << " ";
		cout << endl;

		// arrange
		vector<int> used(n, 0);
		cout << endl;
		cout << arrange(elems, next, used, vector<char>()) << endl;
	}
	return 0;
}