SHUOJ 241 Franklin's Trouble（简单数学题）

http://acmoj.shu.edu.cn/openjudge/viewproblem?coll_id=1&prob_id=241

 

 

 

Franklin's Trouble

Description

Professor Franklin is consulting for an oil company, which is planning a large pipeline running east to west through an oil field of M wells. From each well, a sub-pipeline is to be connected directly to the main pipeline along a shortest path (either north or south). Given x- and y-coordinates of the wells, the professor wants to pick the optimal location of the main pipeline, which means to minimize the total length of the sub-pipelines. Franklin is not good at calculation, but you are. Can you help him?

Input

The first line of input contains a single integer N representing the number of oil fields. Then N field descriptions follow. Each field description starts with an integer M representing the number of wells in this field. Each well is represented by a point (x, y) (both x and y are integers). You can assume all integers are between 1 and 100 and no two wells share the same x-coordinate.

Output

For each field, output the total length of the sub-pipelines. The length should be minimized.

Sample Input

1

2

1 0

2 1

 

Sample Output

1

 

Explanation

In sample input, there is only one case. In this case, the main pipeline can be located at any position between y=0 and y=1 lines to reach the optimal result 1.

 

 

 

题目意思就是M个点，求一条水平线，使得所有点到水平线的距离之和最小。输出最小距离之和

输入数据的X坐标不相等。

 

很简单的数学题，明显应该选在Y坐标的中位数处，可以使得距离之和最小；；

 

#include<stdio.h>

#include<algorithm>

using namespace std;

const int MAXN=1000;

int y[MAXN];

int main()

{

    int T;

    int n;

    int x;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        for(int i=0;i<n;i++)

          scanf("%d%d",&x,&y[i]);

        sort(y,y+n);

        int tmp=y[n/2];

        int res=0;

        for(int i=0;i<n;i++)

          res+=abs(tmp-y[i]);

        printf("%d\n",res);

    }    

    return 0;

}