PAT甲级1033

1033. To Fill or Not to Fill (25)

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

ZHANG, Guochuan

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: C_max (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; D_avg (<=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

#include
#include
using namespace std;
const int maxn = 510;
const int INF = 1000000000;
struct station
{
	double price, dis;//价格、与起点的距离
}st[maxn];
bool cmp(station a, station b)
{
	return a.dis < b.dis;//按距离从小到大
}
int main()
{
	int n;
	double Cmax, D, Davg;
	scanf("%lf%lf%lf%d", &Cmax, &D, &Davg, &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%lf%lf", &st[i].price, &st[i].dis);
	}
	st[n].price = 0;//数组最后面放置终点，价格为0
	st[n].dis = D;//终点距离为D
	sort(st, st + n, cmp);//将所有加油站按距离从小到大排序
	if (st[0].dis != 0)//如果排序后的第一个加油站距离不是0，说明无法前进
	{
		printf("The maximum travel distance = 0.00\n");
	}
	else
	{
		int now = 0;//当前所处的加油站编号
		//总花费、当前油量及满油行驶距离
		double ans = 0, nowTank = 0, MAX = Cmax*Davg;
		while (now < n)//每次循环选出下一个需要加油的站
		{//选出从当前加油站满油能到达范围内的第一个油价低于当前油价的加油站
			//如果没有低于当前油价的加油站，则选择价格最低的那个
			int k = -1;//最低油价的加油站的编号
			double priceMin = INF;//最低油价
			for (int i = now + 1; i <= n&&st[i].dis - st[now].dis <= MAX; i++)
			{
				if (st[i].price < priceMin)//如果油价比当前最低油价低
				{
					priceMin = st[i].price;//更新最低油价
					k = i;
					//如果找到第一个油价低于当前油价的加油站，直接中断循环
					if (priceMin < st[now].price)
					{
						break;
					}
				}
			}
			if (k == -1)break;//满油状态下无法找到加油站，退出循环输出的结果
			//下面为能找到可到达的加油站k，计算转移花费
			//need为从now到k需要的油量
			double need = (st[k].dis - st[now].dis) / Davg;
			if (priceMin < st[now].price)//如果加油站k的油价低于当前油价
			{
				//只买足够到达加油站k的油
				if (nowTank < need) {//如果当前油量不足need
					ans += (need - nowTank)*st[now].price;//补足need
					nowTank = 0;//到达加油站k后油箱内油量为0

				}
				else//如果当前油量超过need
				{
					nowTank -= need;//直接到达加油站k	
				}
			}
			else//如果加油站k的油价高于当前油价
			{
				ans += (Cmax - nowTank)*st[now].price;//将油箱加满
				//到达加油站k后油箱内油量为Cmax-need
				nowTank = Cmax - need;
			}
			now = k;//到达加油站k，进入下一层循环
		}
		if (now == n)//能到达终点
		{
			printf("%.2f\n", ans);
		}
		else//不能到达终点
		{
			printf("The maximum travel distance = %.2f\n", st[now].dis + MAX);
		}
	}
	return 0;
}