PATA 1033 To Fill or Not to Fill（25 分）解题报告

1033 To Fill or Not to Fill（25 分）

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 = Xwhere 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

作者: ZHANG, Guochuan

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

AC代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct station{
	double price;
	double dis;
}sta[505];
bool cmp(station a,station b);
int main()
{
	double cmax,distance,davg;
	int n;
	cin>>cmax>>distance>>davg>>n;
	for(int i=0;i<n;i++)
		cin>>sta[i].price>>sta[i].dis;
	sta[n].price=0;
	sta[n].dis=distance;
	sort(sta,sta+n,cmp);
	if(sta[0].dis!=0)//出发点没有加油站，走不了 
		cout<<"The maximum travel distance = 0.00";
	else
	{
		int now=0;//now是实时站点 
		double prices=0,nowtank=0;//nowtank是油箱目前油量，prices是总花费 
		double range=cmax*davg;//续航里程 
		while(now<n)//达到目的地站点n之前一直循环 
		{
			int k=-1;
			double Min_price=1000000000;
			for(int i=now+1;i<=n&&sta[i].dis-sta[now].dis<=range;i++)//在续航里程内检索加油站 
			{
				if(sta[i].price<Min_price)//寻找里程内油价最低的加油站 
				{
					Min_price=sta[i].price;
					k=i;
					if(Min_price<sta[now].price)
					//找到比当前加油站更便宜的加油站，立即退出循环，前往该加油站 
						break; 
				}
			}
			if(k==-1)//没有符合条件的加油站，退出循环 
				break;
			double need=(sta[k].dis-sta[now].dis)/davg; 
			if(Min_price<sta[now].price)//里程范围里有比现在加油站便宜的油，直接进到目的地 
			{
				if(nowtank<need)//如果油箱的油不够达到目的地，补上恰好能到目的地的油 
				{
					prices+=(need-nowtank)*sta[now].price;
					nowtank=0;
				}
				else//油箱的油足够到达目的地 
					nowtank-=need;
			}
			else//接下来里程内没有比现在便宜的加油站，在当前的加油站把油箱加满(先加满最便宜的油) 
			{
				prices+=(cmax-nowtank)*sta[now].price;
				nowtank=cmax-need;
			}
			now=k;//更新到目的地加油站 
		}
		if(now==n)
			printf("%.2f",prices);
		else//没有到达目的地，输出能达到的最大距离，即现在所在加油站和出发点的距离，和续航里程range 
			printf("The maximum travel distance = %.2f",sta[now].dis+range);
	}
	return 0;
}

bool cmp(station a,station b)
{
	return a.dis<b.dis;
}
//我们把目的地看做一个油价为0，距离为distance的加油站，放到加油站数组的最后 
//1.算法的核心思想是，如果在能达到的范围内有比现在油箱里的油更便宜的加油站
//2.那就先加恰好能到那个加油站的油，到达目的加油站之后，再重复1st
//3.如果在能抵达的范围内没有比油箱里的油更便宜的，就找范围内最便宜的加油站直接抵达
//然后再站在新的站点上考虑里程范围内，重复1st