• POJ 2540 Hotter Colder --半平面交


    题意: 一个(0,0)到(10,10)的矩形,目标点不定,从(0,0)开始走,如果走到新一点是"Hotter",那么意思是离目标点近了,如果是"Colder“,那么就是远了,"Same"是相同。要你推测目标点的可能位置的面积。

    解法:半平面交水题。从一个点到另一个点远了,说明目标点在两点之间连线的中垂线的离源点较近的一侧,即我们每次都可以得到一条直线来切割平面,要么切割左侧,要么切割右侧,要么都切,再求一个半平面交就可以得出可能面积了。

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #define pi acos(-1.0)
    #define eps 1e-8
    using namespace std;
    
    struct Point{
        double x,y;
        Point(double x=0, double y=0):x(x),y(y) {}
        void input() { scanf("%lf%lf",&x,&y); }
    };
    typedef Point Vector;
    struct Line{
        Point p;
        Vector v;
        double ang;
        Line(){}
        Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }
        Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }
        bool operator < (const Line &L)const { return ang < L.ang; }
    };
    int dcmp(double x) {
        if(x < -eps) return -1;
        if(x > eps) return 1;
        return 0;
    }
    template <class T> T sqr(T x) { return x * x;}
    Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
    Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
    Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
    Vector operator / (Vector A, double p) { return Vector(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); }
    bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
    bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
    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; }
    double Length(Vector A) { return sqrt(Dot(A, A)); }
    double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
    double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
    Vector VectorUnit(Vector x){ return x / Length(x);}
    Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
    double angle(Vector v) { return atan2(v.y, v.x); }
    
    Point GetLineIntersection(Line A, Line B) {
        Vector u = A.p - B.p;
        double t = Cross(B.v, u) / Cross(A.v, B.v);
        return A.p + A.v*t;
    }
    double DisP(Point A,Point B) {
        return Length(B-A);
    }
    double CalcConvexArea(Point* p,int n) {        //凸包面积
        double area = 0.0;
        for(int i=1;i<n-1;i++)
            area += Cross(p[i]-p[0],p[i+1]-p[0]);
        return fabs(area*0.5);
    }
    bool OnLeft(Line L, Point p) { return dcmp(Cross(L.v,p-L.p)) > 0; }
    bool CmpPolarLine(Line a,Line b) {        //直线极角排序
        return angle(a.v) < angle(b.v);
    }
    int HalfPlaneIntersection(Line* L, int n, Point* poly) {    //半平面交点存入poly
        sort(L,L+n,CmpPolarLine);
        int first,last;
        Point *p = new Point[n];
        Line  *q = new Line[n];
        q[first=last=0] = L[0];
        for(int i=1;i<n;i++) {
            while(first < last && !OnLeft(L[i],p[last-1])) last--;
            while(first < last && !OnLeft(L[i],p[first]))  first++;
            q[++last] = L[i];
            if(dcmp(Cross(q[last].v, q[last-1].v)) == 0) {
                last--;
                if(OnLeft(q[last], L[i].p)) q[last] = L[i];
            }
            if(first < last) p[last-1] = GetLineIntersection(q[last-1],q[last]);
        }
        while(first < last && !OnLeft(q[first],p[last-1])) last--;
        if(last-first <= 1) return 0;       //点或线或无界平面,返回0
        p[last] = GetLineIntersection(q[last],q[first]);
        int m = 0;
        for(int i=first;i<=last;i++) poly[m++] = p[i];
        delete p; delete q;
        return m;
    }
    
    Line L[102],TL[103];
    Point poly[104];
    
    int main()
    {
        int i,j,tot = -1;
        Point n,p;
        char ss[10];
        p.x = p.y = 0.0;
        TL[++tot] = Line(Point(0,0),Vector(10,0));
        TL[++tot] = Line(Point(10,0),Vector(0,10));
        TL[++tot] = Line(Point(10,10),Vector(-10,0));
        TL[++tot] = Line(Point(0,10),Vector(0,-10));
        while(scanf("%lf%lf%s",&n.x,&n.y,ss)!=EOF)
        {
            if(ss[0] == 'H')
                TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(p-n)));
            else if(ss[0] == 'C')
                TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(n-p)));
            else {
                TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(p-n)));
                TL[++tot] = Line(Point((n.x+p.x)/2.0,(n.y+p.y)/2.0),Vector(Normal(n-p)));
            }
            p = n;
            for(i=0;i<=tot;i++) L[i] = TL[i];
            int m = HalfPlaneIntersection(L,tot+1,poly);
            if(!m) puts("0.00");
            else   printf("%.2f
    ",CalcConvexArea(poly,m));
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/whatbeg/p/4198345.html
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