PAT 1033 To Fill or Not to Fill[dp]

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 = 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

 题目大意：第一行包含4个整数，Cmax即汽缸最大容量（<=100），D杭州到终点的距离（<=30000），Davg每单位汽油可以跑多少公里（<=20），N（<=500）加油站总数。

下面是N行，Pi表示当前加油站单位汽油价格，Di表示这个加油站距离杭州（起点）的距离。

求解出一种最省钱的方法从杭州到大目的地。

//那么先按从近到远的排个序，要加就得加满吗?

代码转自：https://www.cnblogs.com/chenxiwenruo/p/6735382.html

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cmath>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=505;
int capacity,dis,davg;
int n;
struct Gas{
    double price;
    int dis;//到起点的距离。
    bool operator<(const Gas tmp)const{//重载运算符。
        return dis<tmp.dis;//当调用sort函数时，就会根据这个来排序。
    }
}gas[maxn];
int main()
{
    scanf("%d %d %d %d",&capacity,&dis,&davg,&n);
    for(int i=0;i<n;i++){
        scanf("%lf %d",&gas[i].price,&gas[i].dis);
    }
    sort(gas,gas+n);
    //最后一个设置为目的地，价格为0
    gas[n].price=0;
    gas[n].dis=dis;
    int maxdrive=capacity*davg; //能开的最大距离
    int driveDis;
    int leftdrive=0; //跑到下一个加油站后，还能再跑多少的距离
    int now=0; //车所在的距离
    double sum=0;
    for(int i=0;i<n;i++){
        if(now==gas[i].dis)
            driveDis=maxdrive;
        else
            break; //表明一开始出发的地方就没有加油站
        double minprice=INF;
        int minid=-1;
        for(int j=i+1;j<=n;j++){
            if(now+driveDis>=gas[j].dis){
                //若找到第一个比当前车站价格小的车站，就不继续往下找了
                if(gas[j].price<gas[i].price){
                    minid=j;
                    minprice=gas[j].price;
                    break;
                }
                //否则，就一直找出其中费用最小的
                if(gas[j].price<minprice){
                    minprice=gas[j].dis;
                    minid=j;
                }
            }
        }
        if(minid==-1){
            now=gas[i].dis+maxdrive;
            break;
        }
        else{
            if(minprice<gas[i].price){
    //如果有费用比当前车站i更小的车站，那么加油量只需让车能够达到车站即可
                sum+=((gas[minid].dis-gas[i].dis)*1.0/davg-leftdrive*1.0/davg)*gas[i].price;
                leftdrive=0;
                i=minid-1;//这样下次停在这里。
            }
            else{
    //如果没有费用更小的，那么在i处把油加满，达到后面费用最小的那个车站
                sum+=(capacity-leftdrive*1.0/davg)*gas[i].price;
                leftdrive=maxdrive-(gas[minid].dis-gas[i].dis);
                i=minid-1;
            }
            now=gas[minid].dis;
        }
    }
    if(now==dis){
        printf("%.2lf
",sum);
    }
    else{
        printf("The maximum travel distance = %.2lf
",(double)now);
    }
    return 0;
}

//这个真的很厉害，要多复习！