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 ", 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; }