PTA(Advanced Level)1033.To Fill or Not to Fill

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: Cma**x (≤ 100), the maximum capacity of the tank; D (≤30000), the distance between Hangzhou and the destination city; Dav**g (≤20), the average distance per unit gas that the car can run; and N (≤ 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: P**i, the unit gas price, and D**i (≤D), the distance between this station and Hangzhou, for i=1,⋯,N. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print The maximum travel distance = X where X is the maximum possible distance the car can run, accurate up to 2 decimal places.

Sample Input 1:

50 1300 12 8
6.00 1250
7.00 600
7.00 150
7.10 0
7.20 200
7.50 400
7.30 1000
6.85 300

Sample Output 1:

749.17

Sample Input 2:

50 1300 12 2
7.10 0
7.00 600

Sample Output 2:

The maximum travel distance = 1200.00

思路

• 贪心策略
  • 优先前往更低油价的加油站(\(K\))：加到刚好能到\(K\)的油量，加更便宜的油
  • 没有更低油价的：在当前油站处加满
  • 没有加油站可以到达的时候：到极限了
• 每到新的一个加油站都要根据上面的策略做出决策

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
	double pi, di;   //价格和到起点的距离
}a[510];
bool cmp(node a, node b)
{
	return a.di < b.di;
}

int main()
{
	double capacity, distance, run_per_gas;
	int stations;
	cin >> capacity >> distance >> run_per_gas >> stations;
	for(int i=0;i<stations;i++)
		cin >> a[i].pi >> a[i].di;
	a[stations].pi = 0;
	a[stations].di = distance;   //将重点堪称油价为0，距离起点为D的加油站
	sort(a, a+stations, cmp);

	if(a[0].di != 0)    //一开始的油量为0，如果最近的加油站距离不是0那么肯定不行
	{
		cout << "The maximum travel distance = 0.00\n";
		return 0;
	}
	int pos = 0;    //当前所在的加油站
	double best_len = capacity * run_per_gas;    //满油状态下能走得最远的距离
	double gas_owned = 0.0;    //当前状态之下有的油量
	double cheapest = 0.0;     //最后要花费的油钱
	while(pos < stations)
	{
		int next = -1;
		double min_price = 2100000000;
		for(int i=pos+1;i<=stations && a[i].di - a[pos].di <= best_len;i++)    //当前油量能走的最远距离
		{
			if(a[i].pi < min_price)
			{
				min_price = a[i].pi;
				next = i;
				if(min_price < a[pos].pi)
					break;
			}
		}
		if(next == -1)		break;

		double need = (a[next].di - a[pos].di) / run_per_gas;   //所需油量
		if(min_price < a[pos].pi)   //加油站k的油价小于当前油价
		{
			if(gas_owned < need)    //油量不足
			{
				cheapest += (need - gas_owned) * a[pos].pi;  //补足然后去下个更便宜的加油站
				gas_owned = 0;
			}else	gas_owned -= need;
		}else
		{
			cheapest += (capacity - gas_owned) * a[pos].pi;
			gas_owned = capacity - need;
		}
		pos = next;
	}
	if(pos == stations)
		printf("%.2f\n", cheapest);
	else
		printf("The maximum travel distance = %.2f\n", a[pos].di + best_len);
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805458722734080