1033. To Fill or Not to Fill (25)-贪心（难）

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
注：该贪心算法思想：设满箱油时，车辆最远行驶距离为maxd
按距离排序后，最初处于起点加油站记为now，在距离该站maxd范围内，找出第一个比该站油价更低的站k，到k站加油，若找不到比now站低的站，就找距离该站maxd范围内now站除外油价最低的站k，到该站加油，更新now为k

#include<cstdio>#include<algorithm>using namespace std;const int maxn=510;const int INF=1000000;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;    st[n].dis=D;    sort(st,st+n,cmp);    if(st[0].dis!=0){        printf("The maximum travel distance = 0.00\n");    }else{        int now=0;//当前所处加油站编号         double ans=0,nowTank=0,MAX=Cmax*Davg;//ans是到达当前加油站时的最低花费，nowTank是到达当前加油站时油箱的油量         while(now<n){//每次循环将选出下一个需要到达的加油站             int k=-1;//选出的加油站编号             double priceMin=INF;            for(int i=now+1;i<=n&&st[i].dis-st[now].dis<=MAX;i++){//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                 if(st[i].price<priceMin){                    priceMin=st[i].price;                    k=i;                    if(priceMin<st[now].price){                        break;                    }                }            }            if(k==-1) break;            double need=(st[k].dis-st[now].dis)/Davg;            if(priceMin<st[now].price){                if(nowTank<need){                    ans+=(need-nowTank)*st[now].price;                    nowTank=0;//到达下一个加油站，更新油箱中的油量                 }else{                    nowTank-=need;//到达下一个加油站，更新油箱中的油量                }            }else{                ans+=(Cmax-nowTank)*st[now].price;                nowTank=Cmax-need;//到达下一个加油站，更新油箱中的油量            }            now=k;//到下一个加油站         }        if(now==n){//能够到达终点站             printf("%.2f\n",ans);        }else{            printf("The maximum travel distance = %.2f\n",st[now].dis+MAX);        }    }    return 0;}