871. 最低加油次数

871. 最低加油次数

解题思路:在startFuel条件下能开到最远的加油站limit,在0-limit个加油站中找到油量最大的加油站加油,重复这个操作,如果limit==-1代表能跑到的加油站里所有的油都被加光也没能到达目的地;

class Solution {
public:
	set<int>stations_empty;
	int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations)
	{
		//dis(stations);
		int len = 0;
		if (startFuel>=target)
		{
			return 0;
		}
		if (stations.empty()&&startFuel<target)
		{
			return -1;
		}
		if (startFuel<stations[0][0])
		{
			return -1;
		}
		int limit = -1;
		while (startFuel<target)
		{
			limit = max_len(startFuel, stations);//汽车在只消耗初始燃油状态下能开到最远的加油站
			if (stations.empty() || limit == -1)
			{
				break;
			}

			pair<int, int>_fuel = fuel(stations, limit);
			if(stations_empty.find(_fuel.first)==stations_empty.end())
			{
				stations_empty.insert(_fuel.first);
				startFuel+=_fuel.second;
				len++;			
				vector<int> te = { _fuel.first,_fuel.second };
				auto pos = find(stations.begin(), stations.end(), te);
				stations.erase(pos);
				//dis(stations);
			}


		}
		if (startFuel<target||limit==-1)
		{
			return -1;
		}
		return len;
	}
	int max_len(int startFuel, vector<vector<int>>& stations)
	{
		for (int i = 0; i < stations.size(); i++)
		{
			if (startFuel<=stations[i][0])
			{
				if (startFuel == stations[i][0])
				{
					return i;
				}
				else
					return i - 1;
			}
		}
		return stations.size() - 1;
	}
	pair<int,int> fuel(vector<vector<int>> stations,int limit)
	{
		pair<int, int> ans;
		for (int i = 0; i <= limit; i++)
		{
			if (ans.second<stations[i][1])
			{
				ans.first = stations[i][0];
				ans.second = stations[i][1];
			}
		}
		return ans;
	}
	void dis(vector<vector<int>>n)
	{
		system("cls");
		for (int i = 0; i < n.size(); i++)
		{
			cout << n[i][0] << " " << n[i][1] << endl;
		}
	}
};

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转载自blog.csdn.net/kongqingxin12/article/details/83818417