POJ 3632 Optimal Parking

最新推荐文章于 2019-04-06 07:39:59 发布

原创最新推荐文章于 2019-04-06 07:39:59 发布 · 545 阅读

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#HDU #杭电多校联赛

本文介绍了一个简单的算法问题——最优停车策略。问题设定在一个直线状的长街上，给出多个商店的位置坐标，目的是找到一个停车点，使得从该点出发访问所有商店并返回的总行走距离最短。解决方案是确定商店位置的最大值和最小值，两者之差乘以2即为最短总行程。

Optimal Parking

Description

When shopping on Long Street, Michael usually parks his car at some random location, and then walks to the stores he needs. Can you help Michael choose a place to park which minimises the distance he needs to walk on his shopping round? Long Street is a straight line, where all positions are integer. You pay for parking in a specific slot, which is an integer position on Long Street. Michael does not want to pay for more than one parking though. He is very strong, and does not mind carrying all the bags around.

Input

The first line of input gives the number of test cases, 1 ≤ t ≤ 100. There are two lines for each test case. The first gives the number of stores Michael wants to visit, 1 ≤ n ≤ 20, and the second gives their ninteger positions on Long Street, 0 ≤ x_i ≤ 99.

Output

Output for each test case a line with the minimal distance Michael must walk given optimal parking.

Sample Input

2
4
24 13 89 37
6
7 30 41 14 39 42

Sample Output

152
70

这道题真的是很无奈了，很水的一道题。一分钟就能写完，但是因为是个人训练，看了好几遍都没看懂。

后来过了一道题之后回来看这道题才看懂。这道题的意思就是一条街道，给你其中的几个点，问你把车停在哪里可以保证走遍所有点的情况下走的最短的路，输出走的最短距离。直接找到最小位置和最大位置，相减后*2就行了（去了还要返回的）。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<iomanip>
using namespace std;
int main()
{
	int T,n;
	int a;
	cin>>T;
	int mx,mn;
	while(T--)
	{
		cin>>n;
		cin>>a;
		mx=mn=a;
		for(int i=1;i<n;i++)
		{
			cin>>a;
			if(mx<a)
				mx=a;
			if(mn>a)
				mn=a;
		}
		cout<<2*(mx-mn)<<endl;
	}
	return 0;
}