[Explanations] Magic Line- Computational Geometry (2019 cattle off more school third title H)

Topic links: https://ac.nowcoder.com/acm/contest/883/H

 

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Meaning of the questions:

You give an even number of coordinates of points, to identify a line number these points into n number of equal portions of two

And takes two different points on this line, this line represents

 

Ideas:

See the first reaction this problem is to first set a relatively fixed points at infinity

Then take over sweep, one takes a point, one can find these points is divided into m and m + 1 points

Finally, the slope slightly tilted, and then take it on a line to

Finally crazy WA, do not know the problem or algorithm code issues

After listening to explain the big brother, I found myself really stupid

Last practice:

These coordinates x n points by small to large, x coordinates with the same y-coordinate by ascending sort

This allows to find a center-left point p (x0, y0), taking a vertical straight line x = x0

Then these straight lines of vertical, the center point p is a little rotated counterclockwise

The answer is provided two o'clock a (x1, y2), b (x2, y2), so that x1 = x0-1, x2 = x0 + 1, y1 = y0 + 1000000, y2 = y0-1000000 can be achieved

This will be divided into m points and the point m + 1

Finally, a point remains fixed, will move a unit length y2 = y2 + 1 on the point b

That the point p can also fall below a straight line, into an equal number of piles success

Sigh again their own good fishing qaq

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define e 990000000
 5 
 6 struct point{
 7     int x,y;
 8 }p[1006];
 9 
10 bool cmp(point a,point b){
11     if(a.x==b.x)
12         return a.y<b.y;
13     return a.x<b.x; 
14 } 
15 
16 int main()
17 {
18     int n,m,t,x1,x2,y1,y2;
19     cin>>t;
20     while(t--){
21         cin>>n;
22         for(int i=0;i<n;i++)
23             cin>>p[i].x>>p[i].y;
24         sort(p,p+n,cmp);
25         int m=n/2-1;
26         x1=p[m].x-1,x2=p[m].x+1,y1=p[m].y+e,y2=p[m].y-e+1;
27         printf("%d %d %d %d\n",x1,y1,x2,y2);
28     }
29     return 0; 
30 }