uva 12307 - Smallest Enclosing Rectangle （求凸包最小面积外接矩阵、求凸包最小周长外接矩阵、旋转卡壳、模板）

题目：传送门

 

思路： 求面积和求周长是一样的思路。

　　　 求面积的步骤可 戳

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#define LL long long
#define mem(i, j) memset(i, j, sizeof(i))
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define dep(i, j, k) for(int i = k; i >= j; i--)
#define pb push_back
#define make make_pair
#define INF INT_MAX
#define inf LLONG_MAX
#define PI acos(-1)
using namespace std;

const int N = 1e5 + 5;

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数
};

typedef Point Vector;
/// 向量+向量=向量， 点+向量=向量
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
///点-点=向量
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
///向量*数=向量
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
///向量/数=向量
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }

const double eps = 1e-10;
int dcmp(double x) {
    if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

bool operator < (const Point& a, const Point& b) {
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}

bool operator == (const Point& a, const Point &b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } /// 点积
double Length(Vector A) { return sqrt(1.0 * Dot(A, A)); } /// 计算向量长度
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A、B夹角
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } /// 叉积

int ConvexHull(Point* p, int n, Point* ch) {
    sort(p, p + n);
    int m = 0;
    rep(i, 0, n - 1) {
        while(m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    dep(i, 0, n - 2) {
        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--;
    return m;
}

double PTS(Point p, Point A, Point B) { /// 求点 p 到线段 AB 的最短距离
    if(A == B) return Length(p - A);
    Point v1 = B - A, v2 = p - A, v3 = p - B;
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else return fabs(Cross(v1, v2)) / Length(v1);
}
/* 平行线段 a1a2 和 b1b2 的距离 */
double DS(Point a1, Point a2, Point b1, Point b2) {
    return min(PTS(b1, a1, a2), min(PTS(b2, a1, a2), min(PTS(a1, b1, b2), PTS(a2, b1, b2))));
}

/* 得到向量 a1a2 和 b1b2 的位置关系 */
double Get_angle(Point a1, Point a2, Point b1, Point b2) {
    Point t = b1 - (b2 - a1);
    return Cross(a2 - a1, t - a1);
}

void Rotating_calipers(Point P[], int n) {

    double ans1 = 1e99, ans2 = 1e99;
    int t, l, r;
    t = r = 1;

    if(n < 3) { puts("0.00 0.00"); return ; }

    rep(i, 0, n - 1) {

        while(dcmp(Cross(P[i + 1] - P[i], P[t + 1] - P[i]) - Cross(P[i + 1] - P[i], P[t] - P[i])) > 0) ///确定对踵点
            t = (t + 1) % n;
        while(dcmp(Dot(P[i + 1] - P[i], P[r + 1] - P[i]) - Dot(P[i + 1] - P[i], P[r] - P[i])) > 0) ///确定离 p[i] 最右的那个点
            r = (r + 1) % n;
        if(i == 0) l = r;
        while(dcmp(Dot(P[i + 1] - P[i], P[l + 1] - P[i]) - Dot(P[i + 1] - P[i], P[l] - P[i])) <= 0)///离p[i]最左的那个点
            l = (l + 1) % n;

        double Len = Length(P[i + 1] - P[i]);
        /// 通过p[i]和p[i+1]和p[t](对踵点)确定的三角形的面积的两倍除以底边p[i]p[i+1]就是高了，这个高就是矩形的长
        ///宽度，可以通过最右边那个点在p[i]p[i+1]的投影加上最左边那个点在p[i]p[i+1]的投影就是宽了，代码写的是减，因为一正一负

        double h = Cross(P[i + 1] - P[i], P[t] - P[i]) / Len;
        double w = (Dot(P[i + 1] - P[i], P[r] - P[i]) - Dot(P[i + 1] - P[i], P[l] - P[i])) / Len;

        ans1 = min(ans1, h * w);
        ans2 = min(ans2, 2 * (h + w));
    }
    printf("%.2f %.2f\n", ans1, ans2);
    return ;
}

Point P[N], Q[N];

int main() {
    int n;
    while(scanf("%d", &n) && n) {

        rep(i, 0, n - 1) scanf("%lf %lf", &P[i].x, &P[i].y);

        n = ConvexHull(P, n, Q);

        Rotating_calipers(Q, n);
    }
    return 0;
}