1033. To Fill or Not to Fill (25)

1. 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: 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

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本题的关键点在于
1、寻找可到达路程内，油价低于当前油价的第一个加油站，加油量等于两站行驶需要油量。
2、若不存在该点，加满油，然后行驶至路程内油价最低点。

#include<iostream>#include<algorithm>using namespace std;#define maxn 10000struct node{    double price,disatance;};bool cmp(node a,node b){    return a.disatance<b.disatance;}node station[505];int main(){    int n;    double c,d,da;    cin>>c>>d>>da>>n;    for(int i=0;i<n;i++){        cin>>station[i].price>>station[i].disatance;    }    station[n].disatance=d;    station[n].price=0;    sort(station,station+n,cmp);    if(station[0].disatance!=0){            printf("The maximum travel distance = 0.00\n");        return 0;    }    double nd=0,nc=0,total=0;    int now=0;    while(now<n){        double low=maxn;        int k=-1;        for(int i=now+1;station[i].disatance<=(station[now].disatance+c*da)&&i<=n;i++){            if(station[i].price<low){                k=i;                 low=station[i].price;                if(low<station[now].price) break;            }        }        if(k==-1) break;        double need=(station[k].disatance-station[now].disatance)/da;//本次行驶需要油量        if(station[k].price<station[now].price){            if(nc<need){                total+=(need-nc)*station[now].price;                nc=0;            }            else{                nc-=need;            }        }        else{            total+=(c-nc)*station[now].price;            nc=c-need;        }        now=k;    }    if(now!=n){        printf("The maximum travel distance = %.2f",station[now].disatance+c*da);    }    else {        printf("%.2f",total);    }    return 0;}