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  • POJ 3384 Feng Shui 半平面交

    题目给出两个圆和一个多边形

    问是否能让两个圆在多边形内。

    并且覆盖的面积最大


    圆的半径为r,我们则让多边形的每条边都往内部退r距离。

    然后求半平面交得出的点集中,最远的两个点则是两圆的圆心即可


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <algorithm>
    #include <cstdlib>
    #include <cmath>
    #include <map>
    #include <sstream>
    #include <queue>
    #include <vector>
    #define MAXN 111111
    #define MAXM 211111
    #define PI acos(-1.0)
    #define eps 1e-8
    #define INF 1000000001
    using namespace std;
    int dblcmp(double d)
    {
        if (fabs(d) < eps) return 0;
        return d > eps ? 1 : -1;
    }
    struct point
    {
        double x, y;
        point(){}
        point(double _x, double _y):
        x(_x), y(_y){};
        void input()
        {
            scanf("%lf%lf",&x, &y);
        }
        double dot(point p)
        {
            return x * p.x + y * p.y;
        }
        double distance(point p)
        {
            return hypot(x - p.x, y - p.y);
        }
        point sub(point p)
        {
            return point(x - p.x, y - p.y);
        }
        double det(point p)
        {
            return x * p.y - y * p.x;
        }
        bool operator == (point a)const
        {
            return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
        }
        bool operator < (point a)const
        {
            return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
        }
    
    }p[MAXN];
    struct line
    {
        point a,b;
        line(){}
        line(point _a,point _b)
        {
            a=_a;
            b=_b;
        }
        bool parallel(line v)
        {
            return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0;
        }
        point crosspoint(line v)
        {
            double a1 = v.b.sub(v.a).det(a.sub(v.a));
            double a2 = v.b.sub(v.a).det(b.sub(v.a));
            return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));
        }
        bool operator == (line v)const
        {
        	return (a == v.a) && (b == v.b);
        }
    };
    struct halfplane:public line
    {
    	double angle;
    	halfplane(){}
    	//表示向量 a->b逆时针(左侧)的半平面
    	halfplane(point _a, point _b)
    	{
    		a = _a;
    		b = _b;
    	}
    	halfplane(line v)
    	{
    		a = v.a;
    		b = v.b;
    	}
    	void calcangle()
    	{
    		angle = atan2(b.y - a.y, b.x - a.x);
    	}
    	bool operator <(const halfplane &b)const
    	{
    		return angle < b.angle;
    	}
    };
    struct polygon
    {
        int n;
        point p[MAXN];
        line l[MAXN];
        double area;
        void getline()
        {
            for (int i = 0; i < n; i++)
            {
                l[i] = line(p[i], p[(i + 1) % n]);
            }
        }
        void getarea()
        {
            area = 0;
            int a = 1, b = 2;
            while(b <= n - 1)
            {
                area += p[a].sub(p[0]).det(p[b].sub(p[0]));
                a++;
                b++;
            }
            area = fabs(area) / 2;
        }
    }convex;
    struct halfplanes
    {
    	int n;
    	halfplane hp[MAXN];
    	point p[MAXN];
    	int que[MAXN];
    	int st, ed;
    	void push(halfplane tmp)
    	{
    		hp[n++] = tmp;
    	}
    	void unique()
    	{
    		int m = 1, i;
    		for (i = 1; i < n;i++)
    		{
    			if (dblcmp(hp[i].angle - hp[i - 1].angle))hp[m++] = hp[i];
    			else if (dblcmp(hp[m - 1].b.sub(hp[m - 1].a).det(hp[i].a.sub(hp[m - 1].a)) > 0))hp[m - 1] = hp[i];
    		}
    		n = m;
    	}
    	bool halfplaneinsert()
    	{
    		int i;
    		for (i = 0; i < n; i++) hp[i].calcangle();
    		sort(hp, hp + n);
    		unique();
    		que[st = 0] = 0;
    		que[ed = 1] = 1;
    		p[1] = hp[0].crosspoint(hp[1]);
    		for (i = 2; i < n; i++)
    		{
    			while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[ed].sub(hp[i].a)))) < 0) ed--;
    			while (st < ed && dblcmp((hp[i].b.sub(hp[i].a).det(p[st + 1].sub(hp[i].a)))) < 0) st++;
    			que[++ed] = i;
    			if (hp[i].parallel(hp[que[ed - 1]])) return false;
    			p[ed] = hp[i].crosspoint(hp[que[ed - 1]]);
    		}
    		while (st < ed && dblcmp(hp[que[st]].b.sub(hp[que[st]].a).det(p[ed].sub(hp[que[st]].a))) < 0) ed--;
    		while (st < ed && dblcmp(hp[que[ed]].b.sub(hp[que[ed]].a).det(p[st + 1].sub(hp[que[ed]].a))) < 0) st++;
    		if (st + 1 >= ed)return false;
    		return true;
    	}
    	void getconvex(polygon &con)
    	{
    		p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
    		con.n = ed - st + 1;
    		int j = st, i = 0;
    		for (; j <= ed; i++, j++)
    		{
    			con.p[i] = p[j];
    		}
    	}
    }h;
    int T;
    int n;
    line getmove(point a, point b, double mid)
    {
        double x = a.x - b.x;
        double y = a.y - b.y;
        double L = a.distance(b);
        point ta = point(mid * y / L + a.x, a.y - mid * x / L);
        point tb = point(mid * y / L + b.x, b.y - mid * x / L);
        return line(ta, tb);
    }
    double r;
    int main()
    {
        int cas = 0;
        while(scanf("%d%lf", &n, &r) != EOF)
        {
            for(int i = 0; i < n; i++) p[i].input();
            h.n = 0;
    
            for(int i = 0; i < n; i++)
            {
                line tmp = getmove(p[(i + 1) % n], p[i], r);
                h.push(halfplane(tmp));
            }
            h.halfplaneinsert();
            h.getconvex(convex);
            int id1 = 0, id2 = 0;
            double mx = 0;
            for(int i = 0; i < convex.n; i++)
                for(int j = i + 1; j < convex.n; j++)
                {
                    double len = convex.p[i].distance(convex.p[j]);
                    if(dblcmp(len - mx) > 0) id1 = i, id2 = j, mx = len;
                }
            printf("%f %f %f %f
    ", convex.p[id1].x, convex.p[id1].y, convex.p[id2].x, convex.p[id2].y);
        }
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/keanuyaoo/p/3402511.html
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