poj 2187 Beauty Contest , 旋转卡壳求凸包的直径的平方

#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;

struct Point {
int x, y;
Point(int x=0, int y=0):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
// 假设不希望在凸包的边上有输入点，把两个 <= 改成 <
// 注意：输入点集会被改动
vector<Point> ConvexHull(vector<Point>& p) {
// 预处理。删除反复点
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());

int n = p.size();
int m = 0;
vector<Point> ch(n+1);
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--;
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==1) return 0;
if(n==2) return Dist2(p[0], p[1]);
p.push_back(p[0]);
int ans = 0;
for(int u = 0, v = 1; u < n; ++u) {
for(;;) {
int diff = Cross(p[u+1]-p[u], p[v+1]-p[v]);
if(diff<=0) {
ans = max(ans, Dist2(p[u], p[v]));
if(diff==0) ans = max(ans, Dist2(p[u], p[v+1]));
break;
}
v = (v+1) % n;
}
}
return ans;
}
int main() {
int n;
scanf("%d", &n);
vector<Point> P;
for(int i=0; i<n; ++i) {
int x, y;
scanf("%d%d", &x, &y);
P.push_back(Point(x, y));
}
printf("%d\n", diameter2(P));
return 0;
}