PATA 1033 To Fill or Not to Fill

1033 To Fill or Not to Fill 

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
  

待解决，思考中 

这个题目我想了好几天，知道是使用贪心算法但是对于严谨性一直欠考虑。

以下是《算法笔记》AC代码：

  

   

    

     

    
    

     

      #include<cstdio>
     
    
   

    

     

    
    

     

      #include<iostream>
     
    
   

    

     

    
    

     

      #include<cstring>
     
    
   

    

     

    
    

     

      #include<algorithm>
     
    
   

    

     

    
    

     

      using namespace std;
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      const int maxn = 510;
     
    
   

    

     

    
    

     

      const int INF = 1000000000;
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      struct station{
     
    
   

    

     

    
    

     
	double price,dis;//价格，与起点的距离 
     
    
   

    

     

    
    

     

      }st[maxn];
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      bool cmp(station a, station b)
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
	return a.dis<b.dis;//从小到大排序 
     
    
   

    

     

    
    

     

      }
     
    
   

    

     

    
    

     
 
     
    
   

    

     

    
    

     

      int main()
     
    
   

    

     

    
    

     

      {
     
    
   

    

     

    
    

     
	int n;
     
    
   

    

     

    
    

     
	double Cmax,D,Davg;
     
    
   

    

     

    
    

     
	scanf("%lf%lf%lf%d",&Cmax,&D,&Davg,&n);
     
    
   

    

     

    
    

     
	for(int i=0;i<n;i++)
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     
		scanf("%lf%lf",&st[i].price,&st[i].dis);
     
    
   

    

     

    
    

     

      	}
     
    
   

    

     

    
    

     

      	st[n].price = 0;//终点 
     
    
   

    

     

    
    

     

      	st[n].dis = D;//终点距离 
     
    
   

    

     

    
    

     
	sort(st,st+n,cmp);//加油站从小到大排序 
     
    
   

    

     

    
    

     
	if(st[0].dis!=0)//第一个加油站不为0，无法前进 
     
    
   

    

     

    
    

     

      	{
     
    
   

    

     

    
    

     
		printf("The maximum travel distance = 0.00\n");
     
    
   

    

     

    
    

     

      	 } 
     
    
   

    

     

    
    

     
	 else{
     
    
   

    

     

    
    

     
	 	int now = 0;//当前所处的加油站编号
     
    
   

    

     

    
    

     
		 double ans =0, nowTank=0, MAX = Cmax*Davg;
     
    
   

    

     

    
    

     
		 while(now<n) {
     
    
   

    

     

    
    

     
		 	//每次循环选出下一个需要到达的加油站
     
    
   

    

     

    
    

     
			 int k=-1;
     
    
   

    

     

    
    

     
			 double priceMin=INF;
     
    
   

    

     

    
    

     
			 for(int i=now+1;i<=n&&st[i].dis-st[now].dis<=MAX;i++)
     
    
   

    

     

    
    

     

      			 {
     
    
   

    

     

    
    

     
			 	if(st[i].price<priceMin)
     
    
   

    

     

    
    

     

      			 	{
     
    
   

    

     

    
    

     

      			 		priceMin = st[i].price;
     
    
   

    

     

    
    

     

      			 		k = i;
     
    
   

    

     

    
    

     
			 		if(priceMin<st[now].price)
     
    
   

    

     

    
    

     

      			 		{
     
    
   

    

     

    
    

     
			 			break;
     
    
   

    

     

    
    

     

      					 }
     
    
   

    

     

    
    

     

      				 }
     
    
   

    

     

    
    

     

      			  } 
     
    
   

    

     

    
    

     
			  
     
    
   

    

     

    
    

     
		if(k == -1)	break;
     
    
   

    

     

    
    

     
		double need = (st[k].dis-st[now].dis)/Davg;
     
    
   

    

     

    
    

     
		if(priceMin < st[now].price)
     
    
   

    

     

    
    

     

      		{
     
    
   

    

     

    
    

     
			if(nowTank<need)
     
    
   

    

     

    
    

     

      			{
     
    
   

    

     

    
    

     

      				ans+=(need-nowTank)*st[now].price;
     
    
   

    

     

    
    

     

      				nowTank = 0;
     
    
   

    

     

    
    

     

      			}
     
    
   

    

     

    
    

     
			else{
     
    
   

    

     

    
    

     

      				nowTank -= need;
     
    
   

    

     

    
    

     

      			}
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     
		else{
     
    
   

    

     

    
    

     

      			ans+=(Cmax-nowTank)*st[now].price;
     
    
   

    

     

    
    

     

      			nowTank = Cmax - need;
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     

      		now= k;
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     
		if(now == n)
     
    
   

    

     

    
    

     

      		{
     
    
   

    

     

    
    

     
			printf("%.2f\n",ans);
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     
		else{
     
    
   

    

     

    
    

     
			printf("The maximum travel distance = %.2f\n",st[now].dis+MAX);
     
    
   

    

     

    
    

     

      		}
     
    
   

    

     

    
    

     

      		} 
     
    
   

    

     

    
    

     
		return 0;
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: blog.csdn.net，作者：沧夜2021，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/CANGYE0504/article/details/89553546