HDU4312 Meeting point-2)

HDU4312 Meeting point-2)

Question link:

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Given n (n <= 100000) points on the plane coordinate, select one of them to minimize the Chebyshev distance between all points and the current point.

 

Analysis:

Distance: Set a (x1, y1), B (x2, y2); DIS = max (| x1-x2 |, | y1-y2 |) = (| x1-x2 + y1-y2 | + | x1-x2-y1 + y2 |)/2;

We regard the coordinates of point aa as (x1 + y1, x1-y1), bb as (x2 + y2, x2-y2), which is equivalent

Rotate 45 degrees counterclockwise in the coordinate system, and increase the coordinates by 2 times.

Then, seek the newest half of the distance between the smallest Manhattan.

 

The Code is as follows:

 

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include using namespace std;typedef long long LL;const int maxn = 1e5+10;struct point{    int x,y;    LL sum;}p[maxn];bool cmp1(point A,point B){    if(A.x
     
      = 1; --i) {            p[i].sum += sum - (n-i) * p[i].x;            sum += p[i].x;        }        sum = 0;        sort(p+1, p+1+n, cmp2);        for (LL i = 1; i <= n; ++i) {            p[i].sum += (i-1) * p[i].y -sum;            sum += p[i].y;        }        sum = 0;        LL ans = 1LL<<62;        for (LL i = n; i >= 1; --i) {            p[i].sum += sum - (n-i) * p[i].y;            ans = min(ans, p[i].sum);            sum += p[i].y;        }        printf(%I64d,ans/2);    }    return 0;}
     
    
   
  
 

 

 

 

 

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