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; }