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  • uva 12307

    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

    #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;
    }
    一步一步,永不停息
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  • 原文地址:https://www.cnblogs.com/Willems/p/12511845.html
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