7-7 To Fill or Not to Fill （30 分）

最新推荐文章于 2021-09-26 23:00:55 发布

砖瓦程序猿

最新推荐文章于 2021-09-26 23:00:55 发布

阅读量1.5k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： PAT 文章标签： PAT

本文链接：https://blog.youkuaiyun.com/qq_40099908/article/details/82844612

PAT 专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个算法，用于解决从杭州出发到任意城市驾驶汽车时，考虑到油箱容量限制和沿途加油站不同油价的情况，如何设计最经济的行车路线。输入包括油箱最大容量、目的地距离、平均油耗和加油站信息，输出是最便宜的总花费或无法到达目的地时的最大可行距离。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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: Cmax (≤ 100), the maximum capacity of the tank; D (≤30000), the distance between Hangzhou and the destination city; Davg (≤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: Pi, the unit gas price, and Di (≤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

/**7-7 To Fill or Not to Fill （30 分）*/
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

typedef struct node{
    double v;
    int d;
}node;

bool cmp(const node &n,const node &m){
    return n.d<m.d;
}
int main(){
    int c,x,y,n;
    double oil=0,money=0;
    scanf("%d %d %d %d",&c,&x,&y,&n);
    int maxd = c*y;
    int k=0;
    node *a = (node*)malloc((n+1)*sizeof(node));
    for(int i=0;i<n;i++){
        scanf("%lf %d",&a[i].v,&a[i].d);
    }
    a[n].v=0,a[n].d=x;
    sort(a,a+n,cmp);
//    for(int i=0;i<n;i++)
//        cout<<a[i].v<<" "<<a[i].d<<endl;
    if(a[0].d!=0){
        printf("The maximum travel distance = 0.00");
        return 0;
    }
    while(1){
		int st=k;
		if(a[st+1].d-a[k].d>maxd){/**如果两站之间的距离大于可以行驶的距离*/
			if(x-a[k].d<=maxd){/**if终点到此时的站的距离<=最大距离,直接行使到终点*/
				money +=(x-a[k].d-oil)*a[k].v;
				printf("%.2f",(money)*1.0/y*1.0);
				return 0;
			}
			else{/**即到不了下一站也到不了终点，加满油行驶最大距离*/
				printf("The maximum travel distance = %.2f",(a[k].d+maxd*1.0));
				return 0;
			}
		}
		while(st<n && a[++st].d-a[k].d<=maxd){/**遍历每一个站，if本站到下一站的距离<=可行驶的最大距离时
                                                ，我们就可以继续判断如何向终点行驶*/
			if(a[st].v<a[k].v){ /**if下一站站价钱比当前的站便宜，那么就过去，但是要根据此时的剩余油量可行驶的距离来判断*/
				if(oil>=(a[st].d-a[k].d))/**if此时剩余油量可行驶的距离 >= 两站之间的距离就过去*/
				oil-=a[st].d-a[k].d;
				else{/**如果此时剩余油量可行驶的距离 < 两站之间的距离就不过去*/
				money+=(a[st].d-a[k].d-oil)*a[k].v; /**那么肯定要花钱加油了，加的油量时正好到下一个站的*/
				oil=0;
				}
				k=st;
				break;
			}
		}
		if(a[st].d-a[k].d>maxd){
			money+=(maxd-oil)*a[k].v;
			oil=maxd-(a[st-1].d-a[k].d);
			k=st-1;
        }
	}
	return 0;
}