PAT 1033. To Fill or Not to Fill (贪心)

PAT-A的最后一题，最终做出来了...

是贪心，通过局部最优获得全局最优。

1. 将加油站按距离升序排序

2. 记录当前所在的加油站index，存有的汽油，花费。向后遍历全部 该站可抵达的加油站

3. 遍历中有两种情况：

1) 若发现油价比index更低的站next：

立即跳到该站（此时可能须要加油），不再继续遍历 —— 由于即使想要到达next后面的站，能够通过在next站购买更廉价的汽油来实现

2) 没有发现油价比index更低的站，则选择全部站中油价最低的站作为next:

此时考虑能否通过index抵达终点，若能，直接break, 不用管next;  若不能，则装满油，并行驶到next站.

4. 測试点2測试最大距离为0的情况 —— 即在起始点没有加油站。

代码：

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>

using namespace std;

struct Station
{
	double price;
	int dis;
	Station(double p, int d): price(p), dis(d) {}
	friend bool operator <(const Station& a, const Station& b)
	{
		return a.dis < b.dis;
	}
};

int main()
{
	vector<Station> station;
	double cmax, dest, davg, p;
	int n, d;
	int index = 0; // index indicate the station where they are.
	double cost = 0, gas = 0;
	cin >> cmax >> dest >> davg >> n;
	for (int i = 0; i < n; ++ i)
	{
		cin >> p >> d;
		station.push_back( Station(p, d) );
	}
	sort(station.begin(), station.end());
	if (station[0].dis != 0)
	{
		cout << "The maximum travel distance = 0.00" << endl;
		return 0;
	}

	while ( true )
	{
		// 选择下一个站
		double min_price = 2100000000;
		int next = -1;
		bool find_cheaper = false;
		for (int i = index+1;
			i<n && station[i].dis<dest && station[i].dis<=station[index].dis+cmax*davg;
			++ i)
		{
			if (station[i].price <= station[index].price)
			{
				find_cheaper = true;
				next = i;
				break;
			} else if (station[i].price < min_price)
			{
				min_price = station[i].price;
				next = i;
			}
		}
		// 选择加油方式
		if (find_cheaper == true)
		{
			if (station[next].dis - station[index].dis > gas*davg) // 油无法到达该站
			{
				cost += ((station[next].dis-station[index].dis)/davg - gas) * station[index].price;
				gas = 0;
			} else
			{
				gas -= (station[next].dis-station[index].dis)/davg;
			}
			index = next;
		} else if (next != -1) // 至少能够抵达下一个更贵的站
		{
			if (station[index].dis + cmax*davg >= dest)
			{
				break; // 跳出while循环
			}
			cost = cost + (cmax - gas) * station[index].price; // 装满油
			gas = cmax - (station[next].dis - station[index].dis) / davg;
			index = next;
		} else
		{
			break;
		}
	}

	if (station[index].dis + cmax*davg >= dest)
	{
		cost = cost + ((dest-station[index].dis)/davg-gas)*station[index].price;
		cout << setiosflags(ios::fixed) << setprecision(2) << cost;
	} else
	{
		cout << "The maximum travel distance = " << setiosflags(ios::fixed) << setprecision(2) << station[index].dis + cmax*davg;
	}

	return 0;
}