HDU 1673 Optimal Parking【简单题】

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本文介绍了一个帮助找到最优停车地点的问题解决方法，目的是最小化购物时行走的距离。通过将商店的位置进行排序并找出最大和最小值，可以计算出从任意一点出发来回走两次所需的最短距离。

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/*
中文题目来回走两次
中文翻译-大意去商店购物，商店都是在一条街上面的，所以走完应该是来回走两次的

解题思路：排序找最大值和最小值
关键点：找最小点和最大点
解题人：lingnichong
解题时间：2014-09-21 00:22:09
解题体会：看了很久看懂意思，但方法上用错了，后来想了想，不管停在那里都要将整个街道最短和最长的路走两遍
*/

Optimal Parking

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1719 Accepted Submission(s): 1461

Problem 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 n integer positions on Long Street, 0 <= xi <= 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

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

#include<stdio.h>
#include<algorithm>
using namespace std;

int arr[25];

int main()
{
    int n,m;
    scanf("%d",&n);
    while(n--)
    {
        int i,t,sum=0;
        scanf("%d",&m);
        for(i=0;i<m;i++)
        	scanf("%d",&arr[i]);
    	sort(arr,arr+m);
    	sum=2*(arr[m-1]-arr[0]);
 		printf("%d\n",sum);
    }    
    return 0;
}