1033. To Fill or Not to Fill (25)

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


先按照距离排一下序。然后不断更新最短价格，每次都把油加满，如果遇到价格低的加油站，就在保证能到达当前点的前提下，多余的油清除，价格也相应改变，然后按照当前价格把油加满，如果遇到的价格不低，但是剩下的油不足以到大下一站，也把油加满，更新最小价格。
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define Max 505
double Cmax,D,Dave;
int N,flag = 1;
double capcity = 0,price = 0,dis = 0;
struct tank
{
  double distance,price;
}s[Max];
bool cmp(tank a,tank b)
{
  return a.distance < b.distance;
}
int main()
{
  cin>>Cmax>>D>>Dave>>N;
  for(int i = 0;i < N;i ++)
  {
    cin>>s[i].price>>s[i].distance;
  }
  s[N].distance = D;
  sort(s,s+N+1,cmp);
  double minp;///当前最小价格

  int i;
  for(i = 0;i < N + 1;i ++)
  {
      if(capcity*Dave - (s[i].distance - s[i - 1].distance)  < 0)
        {
            dis = s[i - 1].distance + capcity * Dave;
            flag = 0;
            break;
        }

        capcity -= (s[i].distance - s[i - 1].distance) / Dave;
        if(s[i].distance == D)break;
      if(minp > s[i].price)
        {
            price -= capcity * minp;
            minp = s[i].price;
            price += minp * Cmax;
            capcity = Cmax;
        }
        else if(s[i + 1].distance - s[i].distance > capcity * Dave)
        {
            minp = s[i].price;
            price += (Cmax - capcity) * minp;
            capcity = Cmax;
        }
  }
    if(!flag&&dis < D)printf("The maximum travel distance = %.2f",dis);
    else printf("%.2f",price - capcity * minp);
  return 0;
}