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  • UVA–10652 Board Wrapping[凸包]

     

    题意:

    给出n个矩形的中点、长、宽和顺时针的角度。让你用最小的凸多边形把他们包起来,计算矩形面积占凸多边形的百分比。

    用大白书给出的凸包的算法,将矩形的每个顶点都做一次凸包,求出凸包的面积。

    #include "bits/stdc++.h"
    using namespace std;
    const int maxn = 3000;
    const double eps=1e-8;
    const double PI = acos(-1.0);
    struct Point {
        double x, y;
        Point(double x = 0.0, double y = 0.0):x(x), y(y) {}
    };
    typedef Point Vector;
    Point operator + (Point A, Point B) {
        return Point(A.x+B.x, A.y+B.y);
    }
    Point operator - (Point A, Point B) {
        return Point(A.x-B.x, A.y-B.y);
    }
    Point operator * (Point A, double p) {
        return Point(A.x*p, A.y*p);
    }
    Point operator / (Point A, double p) {
        return Point(A.x/p, A.y/p);
    }
    bool operator < (const Point& a, const Point& b) {
        return a.x<b.x || (a.x==b.x && a.y<b.y);
    }
    int dcmp(double x) {
        if (fabs(x)<eps) return 0;return x<0?-1:1;
    }
    bool operator == (const Point& a, const Point &b) {
        return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y)==0;
    }
    double Dot(Point A, Point B) {
        return A.x*B.x+A.y*B.y;
    }
    double Cross(Point A, Point B) {
        return A.x*B.y - A.y*B.x;
    }
    double Length(Point A) {return sqrt(Dot(A,A));}
    Vector Normal(Vector A) {return Vector(-A.y, A.x)/Length(A);}
    double Angle(Vector A, Vector B) {
        return acos(Dot(A,B)/Length(A)/Length(B));
    } 
    Vector Rotate(Vector A, double rad) {
        return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
    } 
    double PolygonArea(Point* p, int n) {
        double area = 0.0;
        for (int i = 1; i < n-1; i++) 
            area += Cross(p[i]-p[0],p[i+1]-p[0]);
        return area/2.0;
    }
    Point p[maxn], ch[maxn];
    bool cmp(Point a, Point b) {
        if (dcmp(a.x - b.x) == 0) return dcmp(a.y - a.y) <= 0;
        return dcmp(a.x - b.x) < 0;
    }
    //输出n个点,输出ch作为凸包所含的点
    //如果不希望边上有点 将<=改为<   输入无重复点
    int ConvexHull(Point* p, int n, Point* ch) {
        sort(p, p + n); //先比较x,在比较y字典序
        int m = 0;
        for (int i = 0; i < n; i++) {
            while (m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
            ch[m++] = p[i];
        }
        int k = m; 
        for (int i = n - 2; i >= 0; i--) {
            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;
    }
    int main(int argc, char const *argv[])
    {
        int T;
        scanf("%d", &T);
        while (T--) {
            int N, cnt = 0;
            scanf("%d", &N);
            double ara = 0.0;
            for (int  i = 0; i < N; i++) {
                double x, y, w, h, j;
                scanf("%lf%lf%lf%lf%lf", &x, &y, &w, &h, &j);
                double ang = -(j/180.0)*PI;
                Point o = Point(x, y);
                p[cnt++] = o + Rotate(Vector(w/2, h/2), ang);
                p[cnt++] = o + Rotate(Vector(-w/2, h/2), ang);
                p[cnt++] = o + Rotate(Vector(w/2, -h/2), ang);
                p[cnt++] = o + Rotate(Vector(-w/2, -h/2), ang);
                ara += w*h;
            }
            int m = ConvexHull(p, cnt, ch);
            double A = PolygonArea(ch, m);
            printf("%.1lf %%
    ", ara*100/A);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cniwoq/p/7623970.html
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