A1033 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: Cmax (≤ 100), the maximum capacity of the tank; D (≤30000), the distance between Hangzhou and the destination city; Davg (≤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: Pi, the unit gas price, and Di (≤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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
const int maxn=510;
const int INF=100000000;
struct station{
    double price,dis;
}st[maxn];
bool cmp(station a,station b){
    return a.dis<b.dis;
}
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);
    if(st[0].dis!=0)
        printf("The maximum travel distance = 0.00\n");
    else{
        int now=0;//当前所处的加油站编号
        double ans=0,nowTank=0,MAX=Cmax*Davg;
        while(now<n){
            int k=-1;//最低油价的加油站编号
            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){
                    k=i;
                    priceMin=st[i].price;
                    //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;//油满状态，无法找到加油站
            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
                if(nowTank<need){//油不够，需加油
                    ans+=(need-nowTank)*st[now].price;
                    nowTank=0;
                }
                else{//油够
                    nowTank-=need;
                }
            }
            else{//加油站k的油价高于当前油价
                ans+=(Cmax-nowTank)*st[now].price;//给油箱加满
                nowTank=Cmax-need;
            }
            now=k;
        }
        if(now==n)
            printf("%.2f\n",ans);
        else
            printf("The maximum travel distance = %.2f\n",st[now].dis+MAX);
    }
    return 0;
}

来源：CSDN

作者：_APTX4869

链接：https://blog.csdn.net/nanke_4869/article/details/104578910

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4.4 贪心：A1033 To Fill or Not to Fill