学步园

题意：给出点集，求覆盖所有点的最小面积矩形和最小周长矩形。

参考了很多Killerfear前辈的代码：传送门。

枚举每条边为矩形底边，找出最高点和最左最右点。

判断最高点是三角形面积，最左最右点是点积的正负。

最左最右点各占据凸包一半的点。

可以推出这三个点的性质都是单峰函数。

#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
#define MAXN 100010
#define eps 1e-10

//const double pi = acos(-1.0);
inline double sig(double x) { return (x > eps) - (x < -eps); };

typedef struct Point
{
        double x, y;
        Point() {}
        Point(double _x, double _y):
                x(_x), y(_y) {}
        bool operator <(const Point &argu) const { return sig(x - argu.x) == 0 ? y < argu.y : x < argu.x; }
        double dis(const Point &argu) const { return sqrt((x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y)); }
        double dis2(const Point &argu) const { return (x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y); }
        double operator ^(const Point &argu) const { return x * argu.y - y * argu.x; }
        double operator *(const Point &argu) const { return x * argu.x + y * argu.y; }
        Point operator -(const Point &argu) const { return Point(x - argu.x, y - argu.y); }
        double len2() const { return x * x + y * y; }
        double len() const { return sqrt(x * x + y * y); }
        void in() { scanf("%lf%lf", &x, &y); }
        void out() { printf("%.3lf %.3lf\n", x, y); }
}Vector;

inline double Cross(const Point &o, const Point &a, const Point &b) { return (a - o) ^ (b - o); }

int ConvexHull(Point p[], Point ch[], int n)
{
    sort(p, p + n);
    int top = 0;
    for(int i = 0; i < n; i++)
    {
        while(top > 1 && Cross(ch[top - 2], ch[top - 1], p[i]) <= 0) top--;
        ch[top++] = p[i];
    }
    int t = top;
    for(int i = n - 2; i >= 0; i--)
    {
        while(top > t && Cross(ch[top - 2], ch[top - 1], p[i]) <= 0) top--;
        ch[top++] = p[i];
    }
    top--;
    return top;
}

void RotatingCalipers(Point p[], int n, double &mina, double &minp)
{
    int t = 1, l = 1, r = 1;
    mina = minp = 1e15;
    for(int i = 0; i < n; i++)
    {
        //枚举边(p[i], p[i+1])
        while(sig((p[i + 1] - p[i]) ^ (p[t + 1] - p[t])) > 0) t = (t + 1) % n; //找出最高点
        while(sig((p[i + 1] - p[i]) * (p[r + 1] - p[r])) > 0) r = (r + 1) % n; //找出最右点
        if(i == 0) l = (r + 1) % n; //初始化最左点
        while(sig((p[i + 1] - p[i]) * (p[l + 1] - p[l])) < 0) l = (l + 1) % n; //找出最左点
        double d = p[i + 1].dis(p[i]);
        double h = ((p[i + 1] - p[i]) ^ (p[t] - p[i])) / d; //三角形高
        double w = ((p[i + 1] - p[i]) * (p[r] - p[l])) / d; //投影
        mina = min(mina, h * w);
        minp = min(minp, 2 * (h + w));
    }
}

Point pp[MAXN], ch[MAXN];
int n, c;
double mina, minp;

void solve()
{
    c = ConvexHull(pp, ch, n); ch[c] = ch[0];
    RotatingCalipers(ch, c, mina, minp);
    printf("%.2lf %.2lf\n", mina, minp);
}

int main()
{
//    freopen("12307.in", "r", stdin);

    while(scanf("%d", &n) && n)
    {
        for(int i = 0; i < n; i++) pp[i].in();
        solve();
    }
    return 0;
}