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)\n", 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\n", 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\n", ans);
return 0;
}