1033 To Fill or Not to Fill (25分)(贪心)

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​​ (≤), the distance between this station and Hangzhou, for ,. 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
题目分析：一道贪心题 用最少的钱走最多的路径

不会 抄了柳神

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 struct station {
     double price,dist;
 };
 bool compare(const station& a, const station& b)
 {
     return a.dist < b.dist;
 }
 int main()
 {
     double cmax = , d = , davg = ;
     int n = ;
     cin >> cmax >> d >> davg >> n;
     vector<station> s(n + );
     s[] = { 0.0,d };
     double price,dist;
     for (int i = ; i <= n; i++)
     {
         cin >> price >> dist;
         s[i] = { price,dist };
     }
     sort(s.begin(), s.end(), compare);
     double nowdist=, maxdist=,leftdist=,nowprice=, totalprice=;
     if (s[].dist != )
     {
         printf("The maximum travel distance = 0.00");
         return ;
     }
     else
         nowprice = s[].price;
     while (nowdist<d)
     {
         double minprice = INT_MAX, mindist = -;
         maxdist = nowdist + cmax * davg;
         int flag = ;
         for (int i = ; i <= n&&s[i].dist<=maxdist; i++)
         {
             if (s[i].dist <=nowdist)continue;
             if (s[i].price < nowprice)
             {
                 totalprice += (s[i].dist - nowdist - leftdist) * nowprice / davg;
                 leftdist = ;
                 nowdist = s[i].dist;
                 nowprice = s[i].price;
                 flag = ;
                 break;
             }
             if (s[i].price < minprice)
             {
                 minprice = s[i].price;
                 mindist = s[i].dist;
             }
         }
         if (flag ==  && minprice != INT_MAX)
         {
             totalprice += (nowprice * (cmax - leftdist / davg));
             leftdist = cmax * davg - (mindist - nowdist);
             nowdist = mindist;
             nowprice = minprice;
         }
         if (flag ==  && minprice == INT_MAX)
         {
             nowdist += cmax * davg;
             printf("The maximum travel distance = %.2f", nowdist);
             return ;
         }
     }
     printf("%.2f", totalprice);
     return ;
 }