[洛谷P1452]Beauty Contest

题目大意：给你$n$个点，求出其中最远点的距离

题解：求出凸包，最远点一定都在凸包上，可以对每条边求出最远的点（可以双指针），然后求出和这条边的端点的距离，更新答案

卡点：最开始对每个点求出最远点，但这样并不可以双指针怎么搞。

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 50010
int __X, __Y;
inline int sqr(int x) {return x * x;}
struct Point {
	int x, y;
	inline Point() {}
	inline Point(int __x, int __y) : x(__x), y(__y) {}
	inline int abs2() {return sqr(x) + sqr(y);}
	inline friend Point operator - (const Point &lhs, const Point &rhs) {
		return Point(lhs.x - rhs.x, lhs.y - rhs.y);
	}
	inline friend Point operator + (const Point &lhs, const Point &rhs) {
		return Point(lhs.x + rhs.x, lhs.y + rhs.y);
	}
	inline friend int operator * (const Point &lhs, const Point &rhs) {
		return lhs.x * rhs.y - rhs.x * lhs.y;
	}
	void writeln() {
		printf("(%d, %d)
", x, y);
	}
} O, s[maxn], v[maxn];

inline int abs2(const Point &x) {return sqr(x.x) + sqr(x.y);}
inline int dis2(const Point &lhs, const Point &rhs) {return abs2(lhs - rhs);}
inline int det(const Point O, const Point &lhs, const Point &rhs) {
	return (lhs - O) * (rhs - O);
}
inline bool operator < (const Point &lhs, const Point &rhs) {
	Point X = lhs - O, Y = rhs - O;
	int tmp = X * Y;
	return (tmp > 0) || (!tmp && abs2(X) > abs2(Y));
}

int n, miny, ans, tot;
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &s[i].x, &s[i].y);
		if ((s[i].y < s[miny].y) || (s[i].y == s[miny].y && s[i].x < s[miny].x)) miny = i;
	}
	if (n == 2) {
		printf("%d
", dis2(s[0], s[1]));
		return 0;
	}
	std::swap(s[0], s[miny]);
	O.x = s[0].x, O.y = s[0].y;
	std::sort(s + 1, s + n);
	v[tot++] = s[0], v[tot++] = s[1], v[tot++] = s[2];
	for (int i = 3; i < n; i++) {
		for (Point a = v[tot - 2], b = v[tot - 1]; tot > 2 && det(a, b, s[i]) <= 0; a = v[tot - 2], b = v[tot - 1]) tot--;
		v[tot++] = s[i];
	}
	int now = 1;
	for (int l = 0, r, D, tmp; l < tot; l++) {
		r = (l + 1) % tot;
		D = det(v[l], v[r], v[now]);
		while (D < (tmp = det(v[l], v[r], v[(now + 1) % tot]))) D = tmp, now = (now + 1) % tot;
		ans = std::max(ans, std::max(dis2(v[l], v[now]), dis2(v[r], v[now])));
	}
	printf("%d
", ans);
	return 0;
}