【洛谷 P3187】 [HNOI2007]最小矩形覆盖 （二维凸包，旋转卡壳）

题目链接

嗯，毒瘤题。

首先有一个结论，就是最小矩形一定有条边和凸包重合。脑补一下就好了。

然后枚举凸包的边，用旋转卡壳维护上顶点、左端点、右端点就好了。

上顶点用叉积，叉积越大三角形面积越大，对应的高也就越大。两边的点用点积，点积越大投影越大。

然后就是精度问题。这种实数计算最好不要直接用比较运算符，要用差和\(eps\)的关系来比较，我就是一直卡在这里。还好有爆炸\(OJ\)离线题库提供的数据。。。

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 50010;
const double eps = 1e-8;
struct point{
    double x, y;
    inline double dis(){
    	return sqrt(x * x + y * y);
    }
    inline void print(){
    	if(fabs(x) < 1e-10) x = 0;
    	if(fabs(y) < 1e-10) y = 0;
    	printf("%.5lf %.5lf\n", x, y);
    }
}p[MAXN];
inline double sig(double x){
	return (x > eps) - (x < -eps);
}
int operator == (point a, point b){
	return a.x == b.x && a.y == b.y;
}
point operator * (point a, double b){    // ba
	return (point){ a.x * b, a.y * b };
}
double operator * (point a, point b){    // a x b
    return a.x * b.y - b.x * a.y;
}
double operator / (point a, point b){    // a . b
	return a.x * b.x + a.y * b.y;
}
point operator - (point a, point b){     // a - b
    return (point){ a.x - b.x, a.y - b.y };
}
point operator + (point a, point b){     // a + b
    return (point){ a.x + b.x, a.y + b.y };
}
int cmp(const point a, const point b){
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}
inline int judge(point a, point b, point c){    //Kab > Kac
	return (b.y - a.y) * (c.x - a.x) > (c.y - a.y) * (b.x - a.x);
}
inline double mult(point a, point b, point c){
	return (a - c) * (b - c);
}
inline double calc(point a, point b, point c){
	return (b - a) / (c - a);
}
int n, top, tp;
point st[MAXN], ts[MAXN], Ans[5];
double ans = 1e18, d, a, b, L, R;
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
       scanf("%lf%lf", &p[i].x, &p[i].y);
    sort(p + 1, p + n + 1, cmp);
    for(int i = 1; i <= n; ++i){
       if(p[i] == p[i - 1]) continue;
       while(top > 1 && judge(st[top - 1], st[top], p[i])) --top;
       st[++top] = p[i];
    }
    for(int i = 1; i <= n; ++i){
       if(p[i] == p[i - 1]) continue;
	   while(tp > 1 && !judge(ts[tp - 1], ts[tp], p[i])) --tp;
       ts[++tp] = p[i];
    }
    for(int i = tp - 1; i; --i) st[++top] = ts[i];
    --top;
    int j = 2, k = 2, l = 2;
    for(int i = 1; i <= top; ++i){
    	while(sig(mult(st[i], st[i + 1], st[j]) - mult(st[i], st[i + 1], st[j + 1])) <= 0) if(++j > top) j = 1;
    	while(sig(calc(st[i], st[i + 1], st[k]) - calc(st[i], st[i + 1], st[k + 1])) <= 0) if(++k > top) k = 1;
    	if(i == 1) l = k;
    	while(sig(calc(st[i], st[i + 1], st[l]) - calc(st[i], st[i + 1], st[l + 1])) >= 0) if(++l > top) l = 1;
    	d = (st[i] - st[i + 1]).dis();
    	R = calc(st[i], st[i + 1], st[k]) / d;
		L = calc(st[i], st[i + 1], st[l]) / d;
    	b = fabs(mult(st[i], st[i + 1], st[j]) / d);
    	a = R - L;
    	if(a * b < ans){
    	   ans = a * b;
    	   Ans[1] = st[i] + (st[i + 1] - st[i]) * (R / d);
    	   Ans[2] = Ans[1] + (st[k] - Ans[1]) * (b / (st[k] - Ans[1]).dis());
    	   Ans[3] = Ans[2] + (st[i] - Ans[1]) * (a / R);
    	   Ans[4] = Ans[3] + (Ans[1] - Ans[2]);
        }
    }
    printf("%.5lf\n", ans);
    double Min = 1e18, pos;
    for(int i = 1; i <= 4; ++i)
       if(Ans[i].y < Min)
       	 Min = Ans[i].y, pos = i;
    for(int i = pos; i <= 4; ++i)
       Ans[i].print();
    for(int i = 1; i < pos; ++i)
       Ans[i].print();
    return 0;
}