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.

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 300Sample Output 1:
749.17Sample Input 2:
50 1300 12 27.10 07.00 600Sample Output 2:
The maximum travel distance = 1200.00
#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <map>#include <string>#define Max 2000using namespace std;struct station{	int flag;	double Price,Distance;}S[Max];bool cmp (station a,station b){	return a.Distance<b.Distance;}int main(){	double n,k;	int s;	double m;	scanf("%lf%lf%lf%d",&n,&m,&k,&s);	for(int i=0;i<s;i++)	{		scanf("%lf %lf",&S[i].Price,&S[i].Distance);		S[i].flag=0;	}	int z=s,x=0;	double xd,d=0.00,p=0.00,y=0.00;	sort(S,S+s,cmp);//路線圖	S[s].Distance=m;	S[s].Price=0;	S[s].flag=0;	if(S[0].Distance!=0) printf("The maximum travel distance = 0.00/n");    s=s+1;	for(int i=0;i<s;i++)	{		int ff=-1,fff,v=-1,j;		int max=100000000;				for( j=i+1;j<s;j++)		{			if(S[j].Distance<=S[i].Distance+n*k)//可以達到			{												 if(S[j].Price<max)				{					v=j;					max=S[j].Price;					if(max<S[i].Price)				{										break;				}				}			}			else y=n;		}		if(v!=-1)//可以繼續走下去		{			if(S[v].Price<S[i].Price)//存在比A點價格低的后置點		       {				   if(y<(S[v].Distance-S[i].Distance)/k)//油不夠用的				   {			         p+=((S[v].Distance-S[i].Distance)/k-y)*S[i].Price;//先走到B點  正好沒油			         d=S[j].Distance;			          y=0;				   }				   else //油夠用				   {					   d=S[v].Distance;					   y=y-(S[v].Distance-S[i].Distance)/k;				   }				   i=v-1;		       }			   else 			   {				   p+=(n-y)*S[i].Price;//加滿油				   y=n-(S[v].Distance-S[i].Distance)/k;//剩余油				   d=S[v].Distance;				   i=v-1;			   }		}		else {			d=d+y*k;			break;		}	}	if(d>=m) printf("%.2f/n",p);	else printf("The maximum travel distance = %.2f",d);    system("pause");	return 0;}