PAT 甲级 1033 To Fill or Not to Fill（25 分）

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

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 99999999;
struct station {
    double price, dis;
};
bool cmp1(station a, station b) {
    return a.dis < b.dis;
}
int main() {
//	freopen("C:\\1.txt", "r", stdin);
    double cmax, d, davg;
    int n;
    scanf("%lf%lf%lf%d", &cmax, &d, &davg, &n);
    vector<station> sta(n + 1);
    sta[0] = {0.0, d};
    for(int i = 1; i <= n; i++)
        scanf("%lf%lf", &sta[i].price, &sta[i].dis);
    sort(sta.begin(), sta.end(), cmp1);
    double nowdis = 0.0, maxdis = 0.0, nowprice = 0.0, totalPrice = 0.0, leftdis = 0.0;
    // 现在所在加油站距离起始点的距离，最大可到达距离，现在所在加油站的油价，总价，剩余油量可以行驶的路程 
    if(sta[0].dis != 0) { // 出发点无加油站 
        printf("The maximum travel distance = 0.00");
        return 0;
    } else {
        nowprice = sta[0].price;
    }
    while(nowdis < d) {
        maxdis = nowdis + cmax * davg;
        double minPriceDis = 0, minPrice = inf;
        // 最低可到达距离， 最低价格 
        int flag = 0;
        for(int i = 1; i <= n && sta[i].dis <= maxdis; i++) {
            if(sta[i].dis <= nowdis) continue;
            if(sta[i].price < nowprice) { // 找到比起始点油价低的第一个站点 
                totalPrice += (sta[i].dis - nowdis - leftdis) * nowprice / davg;
                leftdis = 0.0;
                nowprice = sta[i].price;
                nowdis = sta[i].dis;
                flag = 1;
                break;
            }
            if(sta[i].price < minPrice) {  // 寻找起始点可达范围内油价最低的加油站 
                minPrice = sta[i].price;
                minPriceDis = sta[i].dis;
            }
        }
        if(flag == 0 && minPrice != inf) { // 如果可到达范围内油价都比起始点高，在起始站点加满油 
            totalPrice += (nowprice * (cmax - leftdis / davg));
            leftdis = cmax * davg - (minPriceDis - nowdis);
            nowprice = minPrice;
            nowdis = minPriceDis;     
        }
        if(flag == 0 && minPrice == inf) { // 在可到达范围内，无加油站 
            nowdis += cmax * davg;
            printf("The maximum travel distance = %.2f", nowdis);
            return 0;
        }
    }
    printf("%.2f", totalPrice);
    return 0;
}