回溯算法（收费公路重建问题）

一、问题描述

给出一个距离的集合D， 求出在x轴，存在哪些点能够组合成这样的距离集合；
假设第一个点在0处：path[1] = 0;
最后一个点是距离集合中最大的距离：path[N] = max(D);
使用堆或是红黑树存放距离集合D；

二、具体的代码如下

#include 
#include 
#include 

using namespace std;

bool place(vector<int>& path, multiset<int>& D, int N, int left, int right) {
	bool found = false;
	if (D.empty())
		return true;
	multiset<int>::iterator iter = D.end();
	--iter;
	path[right] = *iter;

	
	int i, j;
	for (i = 1; i < left; i++) {
		if ((iter = D.find(path[right] - path[i])) != D.end())
			D.erase(iter);
		else
		{
			for (int k = i - 1; k > 0; k--)
				D.insert(path[right] - path[i]);
			break;
		}
	}

	if (i == left) {
		for (i = right + 1; i <= N; i++) {
			if ((iter = D.find(path[i] - path[right])) != D.end())
				D.erase(iter);
			else {
				for (int k = 1; k < left; k++)
					D.insert(path[right] - path[k]);
				for (int k = i - 1; k > right; k--)
					D.insert(path[k] - path[right]);
				break;
			}
		}

		if (i == N + 1) {
			found = place(path, D, N, left, right - 1);
			if (!found) {
				for (i = 1; i < left; i++)
					D.insert(path[right] - path[i]);
				for (i = right + 1; i <= N; i++)
					D.insert(path[i] - path[right]);
			}
			else
				return true;
		}
		path[left] = path[N] - path[right];
		for (i = 1; i < left; i++) {
			if ((iter = D.find(path[left] - path[i])) != D.end())
				D.erase(iter);
			else {
				for (int k = i - 1; k > 0; k--)
					D.insert(path[left] - path[k]);
				break;
			}
		}
		if (i == left) {
			for (i = right + 1; i <= N; i++) {
				if ((iter = D.find(path[i] - path[left])) != D.end())
					D.erase(iter);
				else {
					for (int k = 1; k < left; k++)
						D.insert(path[left] - path[k]);
					for (int k = i - 1; k > right; k--)
						D.insert(path[k] - path[left]);
					break;
				}
			}
			if (i == N + 1) {
				found = place(path, D, N, left + 1, right);
				if (!found) {
					for (i = 1; i < left; i++)
						D.insert(path[left] - path[i]);
					for (i = right + 1; i <= N; i++)
						D.insert(path[i] - path[left]);
				}
				else
					return true;
			}
		}
		return false;
	}

}


bool turnpike(vector<int>& path, multiset<int> &D, int N) {
	path[1] = 0;
	multiset<int>::iterator iter = D.end();
	path[N] = *(--iter);
	--iter;
	D.erase(path[N]);
	path[N - 1] = *iter;
	D.erase(path[N - 1]);
	if ((iter = D.find(path[N] - path[N - 1])) != D.end()) {
		D.erase(iter);
		return place(path, D, N, 2, N - 2);
	}

	return 0;
}

int main()
{
	multiset<int> D;
	int arr[] = { 1, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 8, 10 };
	int len = sizeof(arr) / sizeof(arr[0]);
	for (int i = 0; i < len; i++)
		D.insert(arr[i]);
	int N = 6;
	vector<int> path(N+1);
	turnpike(path, D, N);
	for (auto x: path)
		cout << x << endl;

	return 0;
}