uvalive 4728 Squares

题意：求所有正方形中两点距离最大值的平方值。

思路：旋转卡壳法。

分别用数组和vector存凸包时，旋转卡壳代码有所不同。

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 using namespace std;
 const double inf=0x3f3f3f3f;
 const int N = ;
 const double eps = *1e-;
 const double PI = acos(-1.0);
 using namespace std;

 struct Point
 {
     int x, y;
     Point(int x=, int y=):x(x),y(y) { }
 };

 typedef Point Vector;

 Vector operator - (const Point& A, const Point& B)
 {
     return Vector(A.x-B.x, A.y-B.y);
 }

 int Cross(const Vector& A, const Vector& B)
 {
     return A.x*B.y - A.y*B.x;
 }

 int Dot(const Vector& A, const Vector& B)
 {
     return A.x*B.x + A.y*B.y;
 }

 int Dist2(const Point& A, const Point& B)
 {
     return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
 }

 bool operator < (const Point& p1, const Point& p2)
 {
     return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
 }

 bool operator == (const Point& p1, const Point& p2)
 {
     return p1.x == p2.x && p1.y == p2.y;
 }

 // 点集凸包
 // 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
 // 注意：输入点集会被修改
 vector<Point> ConvexHull(vector<Point>& p)
 {
     // 预处理，删除重复点
     sort(p.begin(), p.end());
     p.erase(unique(p.begin(), p.end()), p.end());

     int n = p.size();
     int m = ;
     vector<Point> ch(n+);
     for(int i = ; i < n; i++)
     {
         while(m >  && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-; i >= ; i--)
     {
         while(m > k && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     if(n > ) m--;
     ch.resize(m);
     return ch;
 }

 // 返回点集直径的平方
 int diameter2(vector<Point>& points)
 {
     vector<Point> p = ConvexHull(points);
     int n = p.size();
     if(n == ) return ;
     if(n == ) return Dist2(p[], p[]);
     p.push_back(p[]); // 免得取模
     int ans = ;
     for(int u = , v = ; u < n; u++)
     {
         // 一条直线贴住边p[u]-p[u+1]
         for(;;)
         {
             // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转
             // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0
             // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)
             // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0
             int diff = Cross(p[u+]-p[u], p[v+]-p[v]);
             if(diff <= )
             {
                 ans = max(ans, Dist2(p[u], p[v])); // u和v是对踵点
                 if(diff == ) ans = max(ans, Dist2(p[u], p[v+])); // diff == 0时u和v+1也是对踵点
                 break;
             }
             v = (v + ) % n;
         }
     }
     return ans;
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         int n;
         scanf("%d", &n);
         vector<Point> points;
         for(int i = ; i < n; i++)
         {
             int x, y, w;
             scanf("%d%d%d", &x, &y, &w);
             points.push_back(Point(x, y));
             points.push_back(Point(x+w, y));
             points.push_back(Point(x, y+w));
             points.push_back(Point(x+w, y+w));
         }
         printf("%d\n", diameter2(points));
     }
     return ;
 }