校赛 选修课网址 1097: Meeting

1097: Meeting
Time Limit: 2 Sec  Memory Limit: 64 MB
Submit: 10  Solved: 7
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Description
There n peoples, they want to meet for discussing something about NBUT ACM Laboratory, so they need to decide a location where they meeting.
But they are lazy, they want a location where the summary of each one’s moving
distance is least, now you know the n peoples’ position, can you help them to calculate the minimum summary of each one’s moving distance. 
Input
Input starts with an integer T(1 <= T <= 20), denoting the number of test case.
For each test case, first line contains an integer n(1 <= n <= 100000), denoting the number of people.
Next line contains n integers Pi(0 <= Pi <= 109), denoting each one’s position. Note that maybe someone stay at the same place. 
Output
For each test case, print the minimum summary of each one’s moving distance. 
Sample Input
1
5
1 2 3 4 5
Sample Output
6
HINT

You can set the location at 3, so the minmum summary of each one’s moving distance is 6(2+1+0+1+2).

找最短路径  sort一下 相对最大的减去相对最小的  只要重复的线段最小就是答案  最短的一般都是在中间

#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<sstream>
using namespace std;
int main()
{
	long long n,m,i,k,sum,max,min;
	int a[100010];
	ios::sync_with_stdio(false);
	cin>>k;
	while(k--)
	{
		sum=0;
		cin>>n;
		memset(a,0,sizeof(a));
		for(i=0;i<n;i++)
		cin>>a[i];
		sort(a,a+n);
		max=n-1;
		min=0;
		while(max>=min)
		{
			sum+=a[max]-a[min];	
			max--;
			min++;
		}
		cout<<sum<<endl;
	}
	return 0;
}