LA 4728凸包算法-旋转卡壳的直径

/*LA 4728
凸包算法-旋转卡壳的直径
没有其他技巧，可作为模板运用
注意operator< 中精度的处理，不然会出错
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <set>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
#define eps 1e-7
#define maxn 401000
using namespace std;


struct Point
{
    double x,y;
    Point() {}
    Point(LL xx,LL yy)
    {
        x=xx+0.0;
        y=yy+0.0;
    }
    bool operator<(const Point& p) const//注意：按照逆时针旋转
    {
        if (fabs(x-p.x)<eps) return y<p.y;//注意，这里eps一定要加，不然，不能正常排序
        else return x<p.x;
    }
} P1[maxn],P2[maxn];

typedef Point Vector;

bool operator==(Point A,Point B)
{
    if ((fabs(A.x-B.x)<eps) && (fabs(A.y-B.y)<eps)) return true;
    else return false;
}
Vector operator-(Point A,Point B)//表示A指向B
{
    return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(Vector A,double k)
{
    return Vector(A.x*k,A.y*k);
}
Vector operator+(Point A,Point B)//表示A指向B
{
    return Vector(B.x+A.x,B.y+A.y);
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
double Area2(Point A,Point B,Point C)
{
    return Cross(B-A,C-A);
}
//p是原先点的数组，n是个数，ch是凸包的点集
//精度要求高是用dcmp比较
//返回凸包点的个数
int ConvexHull(Point *p, int n, Point* ch)         //求凸包
{
    sort(p, p + n);//先按照x，再按照y
    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;
}
//卡壳预备函数
LL cross( const Point &o, const Point &a, const Point & b ) {
    return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y);
}
LL dist( const Point & a, const Point & b ) {
    Point p = a - b;
    return p.x * p.x + p.y * p.y;
}
//计算凸包直径，输入凸包 ch，顶点个数为 n，按逆时针排列，输出直径的平方
LL rotating_calipers(Point *ch,int n)
{
    LL q=1,ans=0;
    ch[n]=ch[0];
    for(int p=0; p<n; p++)
    {
        while(cross(ch[p],ch[p+1],ch[q+1])>cross(ch[p],ch[p+1],ch[q]))
            q=(q+1)%n;
        ans=max(ans,max(dist(ch[p],ch[q]),dist(ch[p+1],ch[q+1])));
    }
    return ans;
}
int t,n,cnt1,cnt2;
int main()
{
    cin>>t;
    while(t--)
    {
        cnt1=0;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            LL x,y,w;
            cin>>x>>y>>w;
            P1[cnt1++]=Point(x,y);
            P1[cnt1++]=Point(x+w,y);
            P1[cnt1++]=Point(x,y+w);
            P1[cnt1++]=Point(x+w,y+w);
        }
        cnt2=ConvexHull(P1,cnt1,P2);
        cout<<rotating_calipers(P2,cnt2)<<endl;
    }
    return 0;
}