LA 4728 Squares

Original question:
The famous Korean IT company NHN plans to make a digital map of the Earth with help for wireless sensors which spread out I N Rough terrains. Each sensor sends a geographical data to NHN. But, due to the inaccuracy of the sensing devices equipped in the sensors, and NHN only knows a square and which each GE Ographical data happens. Thus a geographical data can is any point in a square region. You is asked to solve some geometric problem, known as diameter problem, on these undetermined points in the squares. A diameter for a set of points in the plane are defined as the maximum (Euclidean) distance among pairs of the points in th E set. The diameter is used as a measurement to estimate the geographical size of the set. NHN wants you to compute the largest diameter of the points chosen from the squares. In other words, given a set of squares in the plane, you have to choose exactly one point from each square so that the DIA M-
Eter for the chosen points is maximized. The sides of the squares is parallel to x-axis or y-axis, and the squares could have different sizes, intersect all other , and share the same corners. For example, if there was six squares as in the "the", then the largest diameter is defined as the distance Between corner points of squares S 1 and
S 4.
Given a set of n squares in the plane, write a program to compute the largest diameter D of the points when a point is Cho Sen from all square, and to output D 2, i.e., the squared value of D.
Input
Your program was to read from standard input. The input consists of T test cases. The number of test
Cases T is given in the first line of the input. The first line of all test case contains an integer, N, the
Number of squares, where 2≤n≤100,000. Each line of the next n lines contains three integers, x, y,
and W, where (x, y) is the coordinate of the left-lower corner of a square and W are the length of a side
of the Square; 0≤x,y≤10,000 and 1≤w≤10,000.
Output
Your program is-to-write to standard output. Print exactly one line for each test case. The line should
contain the integral value D 2, where D was the largest diameter of the points when a point is chosen
From each square.
Sample Input
2
3
0 0 1
1 0 2
0 0 1
6
2 1 2
1 4 2
3 2 3
4 4 4
6 5 1
5 1 3
Sample Output
13
85

English:
give you n squares so you find out the square of the farthest value of all the points that make up the square vertices.

#include <bits/stdc++.h> using namespace std;

const double eps=1e-6;
    struct point {double x, y;
Point (Double x=0,double y=0): X (x), Y (y) {}};

Point ps[400005],chs[400005];

#define Vector Point Double sq_dist (point A,point B) {return (b.x-a.x) * (b.x-a.x) + (B.Y-A.Y) * (B.Y-A.Y);} BOOL operator < (const point &a,const Point &b)//sorted by x coordinates from small to large, if X is the same, then follow y from small to large {return a.x<b.x| |
(A.X==B.X&AMP;&AMP;A.Y&LT;B.Y); Vector operator + (vector a,vector B) {return vector (A.X+B.X,A.Y+B.Y);} Sit punctuation Add vector operator-(vector a,vector B) {return vector (A.X-B.X,A.Y-B.Y);}
    Subtract int dcmp (double x) {if (Fabs (x) <eps) return 0;
else return x<0?-1:1;


} Double Cross (Vector a,vector B) {return a.x*b.y-a.y*b.x;}
    The int convexhull (point *p,int n,point *ch)//p is all points, n the number of all points, the point in ch which records the convex hull, and returns the number of convex packets {sort (p,p+n);
    int m=0;
       for (int i=0;i<n;i++) {while (M>1&&cross (Ch[m-1]-ch[m-2],p[i]-ch[m-2]) <=0) m--; Ch[m++]=p[i];
    } int k=m; 
        for (int i=n-2;i>=0;i--) {while (M>k&&cross (Ch[m-1]-ch[m-2],p[i]-ch[m-2]) <=0) m--;
    Ch[m++]=p[i];
    } if (n>1) m--;
return m;
    } double rotating_calipers (point *ch,int N) {int q=1;
    Double ans=0;
    CH[N]=CH[0]; for (int p=0;p<n;p++) {When (Cross (vector (ch[q+1]-ch[p+1]), Vector (ch[p]-ch[p+1)) >cross (Vector (ch[q]-ch[
        P+1]), Vector (ch[p]-ch[p+1])) q= (q+1)%n;
    Ans=max (Ans,max (Sq_dist (ch[p],ch[q)), Sq_dist (ch[p+1],ch[q+1]));
} return ans;
    } int main () {Ios::sync_with_stdio (false);
    int n,t;
    cin>>t;
        while (t--) {cin>>n;
        Double x,y,w;
        int ind=0;
            for (int i=0;i<n;i++) {cin>>x>>y>>w;
            Ps[ind].x=x,ps[ind].y=y;
            ind++;
            Ps[ind].x=x,ps[ind].y=y+w;
            ind++; ps[ind].x=x+w,ps[ind].y=Y
            ind++;
            Ps[ind].x=x+w,ps[ind].y=y+w;
        ind++;
        } int M=convexhull (PS,IND,CHS);
        Double Ans=rotating_calipers (chs,m);
    cout<< (int) ans<<endl;
} return 0;






















 }

Answer:

The bare rotation Jam example, pay attention to the output result when cast to int type.