1008. Elevator (20)

最新推荐文章于 2021-08-18 12:42:21 发布

Tejnx

最新推荐文章于 2021-08-18 12:42:21 发布

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本文介绍了一个简单的算法问题，即计算电梯完成一系列楼层请求所需的总时间。考虑到电梯上下楼及停留的时间成本，通过逐个处理请求并累加所需时间来得出最终结果。

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The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

题意：输入n个电梯要去的楼层的指令，给你上楼下楼和停顿的时间，求完成所有指令的时间

思路：水题，可以直接边输入边判断，只需要记录一下上一个楼层的值，判断是上楼还是下楼就好，这里我用的是队列，道理一样

#include <cstdio>
#include <algorithm>
int queue[1000];
const int cost_up = 6, cost_down = 4 ,cost_stop = 5;
int n;
int main()
{
	int current = 0;
	int next = 0;
	int q_head = 0;
	int ans = 0;
	scanf("%d", &n);;
	for (int i = 0; i < n; i++)
		scanf("%d", &queue[i]);
	next = queue[q_head++];
	while (q_head <= n)
	{
		if (current < next)//上楼
		{
			ans += (next - current)*cost_up;
		}
		else if(current > next)//下楼
		{
			ans += (current - next)*cost_down;
		}
		ans += cost_stop;
		current = next;
		next = queue[q_head++];
	}
	printf("%d\n", ans);
	return 0;
}