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  • poj 两条线段接雨水

    poj An Easy Problem?! POJ - 2826

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <vector>
    #include <algorithm>
    
    using namespace std;
    // `计算几何模板`
    const double eps = 1e-8;
    const double inf = 1e20;
    const double pi = acos(-1.0);
    const int maxp = 1010;
    //`Compares a double to zero`
    int sgn(double x){
        if(fabs(x) < eps)return 0;
        if(x < 0)return -1;
        else return 1;
    }
    //square of a double
    inline double sqr(double x){return x*x;}
    /*
     * Point
     * Point()               - Empty constructor
     * Point(double _x,double _y)  - constructor
     * input()             - double input
     * output()            - %.2f output
     * operator ==         - compares x and y
     * operator <          - compares first by x, then by y
     * operator -          - return new Point after subtracting curresponging x and y
     * operator ^          - cross product of 2d points
     * operator *          - dot product
     * len()               - gives length from origin
     * len2()              - gives square of length from origin
     * distance(Point p)   - gives distance from p
     * operator + Point b  - returns new Point after adding curresponging x and y
     * operator * double k - returns new Point after multiplieing x and y by k
     * operator / double k - returns new Point after divideing x and y by k
     * rad(Point a,Point b)- returns the angle of Point a and Point b from this Point
     * trunc(double r)     - return Point that if truncated the distance from center to r
     * rotleft()           - returns 90 degree ccw rotated point
     * rotright()          - returns 90 degree cw rotated point
     * rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw
     */
    struct Point{
        int id;
        double x,y;
        Point(){}
        Point(double _x,double _y){
            x = _x;
            y = _y;
        }
        void input(){
            scanf("%lf%lf",&x,&y);
        }
        void output(){
            printf("%.2f %.2f
    ",x,y);
        }
        bool operator == (Point b)const{
            return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
        }
        bool operator < (Point b)const{
            return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
        }
        Point operator -(const Point &b)const{
            return Point(x-b.x,y-b.y);
        }
        //叉积
        double operator ^(const Point &b)const{
            return x*b.y - y*b.x;
        }
        //点积
        double operator *(const Point &b)const{
            return x*b.x + y*b.y;
        }
        //返回长度
        double len(){
            return hypot(x,y);//库函数
        }
        //返回长度的平方
        double len2(){
            return x*x + y*y;
        }
        //返回两点的距离
        double distance(Point p){
            return hypot(x-p.x,y-p.y);
        }
        Point operator +(const Point &b)const{
            return Point(x+b.x,y+b.y);
        }
        Point operator *(const double &k)const{
            return Point(x*k,y*k);
        }
        Point operator /(const double &k)const{
            return Point(x/k,y/k);
        }
        //`计算pa  和  pb 的夹角`
        //`就是求这个点看a,b 所成的夹角`
        //`测试 LightOJ1203`
        double rad(Point a,Point b){
            Point p = *this;
            return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
        }
        //`化为长度为r的向量`
        Point trunc(double r){
            double l = len();
            if(!sgn(l))return *this;
            r /= l;
            return Point(x*r,y*r);
        }
        //`逆时针旋转90度`
        Point rotleft(){
            return Point(-y,x);
        }
        //`顺时针旋转90度`
        Point rotright(){
            return Point(y,-x);
        }
        //`绕着p点逆时针旋转angle`
        Point rotate(Point p,double angle){
            Point v = (*this) - p;
            double c = cos(angle), s = sin(angle);
            return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
        }
    };
    /*
     * Stores two points
     * Line()                         - Empty constructor
     * Line(Point _s,Point _e)        - Line through _s and _e
     * operator ==                    - checks if two points are same
     * Line(Point p,double angle)     - one end p , another end at angle degree
     * Line(double a,double b,double c) - Line of equation ax + by + c = 0
     * input()                        - inputs s and e
     * adjust()                       - orders in such a way that s < e
     * length()                       - distance of se
     * angle()                        - return 0 <= angle < pi
     * relation(Point p)              - 3 if point is on line
     *                                  1 if point on the left of line
     *                                  2 if point on the right of line
     * pointonseg(double p)           - return true if point on segment
     * parallel(Line v)               - return true if they are parallel
     * segcrossseg(Line v)            - returns 0 if does not intersect
     *                                  returns 1 if non-standard intersection
     *                                  returns 2 if intersects
     * linecrossseg(Line v)           - line and seg
     * linecrossline(Line v)          - 0 if parallel
     *                                  1 if coincides
     *                                  2 if intersects
     * crosspoint(Line v)             - returns intersection point
     * dispointtoline(Point p)        - distance from point p to the line
     * dispointtoseg(Point p)         - distance from p to the segment
     * dissegtoseg(Line v)            - distance of two segment
     * lineprog(Point p)              - returns projected point p on se line
     * symmetrypoint(Point p)         - returns reflection point of p over se
     *
     */
    struct Line{
        Point s,e;
        Line(){}
        Line(Point _s,Point _e){
            s = _s;
            e = _e;
        }
        bool operator ==(Line v){
            return (s == v.s)&&(e == v.e);
        }
        //`根据一个点和倾斜角angle确定直线,0<=angle<pi`
        Line(Point p,double angle){
            s = p;
            if(sgn(angle-pi/2) == 0){
                e = (s + Point(0,1));
            }
            else{
                e = (s + Point(1,tan(angle)));
            }
        }
        //ax+by+c=0
        Line(double a,double b,double c){
            if(sgn(a) == 0){
                s = Point(0,-c/b);
                e = Point(1,-c/b);
            }
            else if(sgn(b) == 0){
                s = Point(-c/a,0);
                e = Point(-c/a,1);
            }
            else{
                s = Point(0,-c/b);
                e = Point(1,(-c-a)/b);
            }
        }
        void input(){
            s.input();
            e.input();
        }
        void adjust(){
            if(e < s)swap(s,e);
        }
        //求线段长度
        double length(){
            return s.distance(e);
        }
        //`返回直线倾斜角 0<=angle<pi`
        double angle(){
            double k = atan2(e.y-s.y,e.x-s.x);
            if(sgn(k) < 0)k += pi;
            if(sgn(k-pi) == 0)k -= pi;
            return k;
        }
        //`点和直线关系`
        //`1  在左侧`
        //`2  在右侧`
        //`3  在直线上`
        int relation(Point p){
            int c = sgn((p-s)^(e-s));
            if(c < 0)return 1;
            else if(c > 0)return 2;
            else return 3;
        }
        // 点在线段上的判断
        bool pointonseg(Point p){
            return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
        }
        //`两向量平行(对应直线平行或重合)`
        bool parallel(Line v){
            return sgn((e-s)^(v.e-v.s)) == 0;
        }
        //`两线段相交判断`
        //`2 规范相交`
        //`1 非规范相交`
        //`0 不相交`
        int segcrossseg(Line v){
            int d1 = sgn((e-s)^(v.s-s));
            int d2 = sgn((e-s)^(v.e-s));
            int d3 = sgn((v.e-v.s)^(s-v.s));
            int d4 = sgn((v.e-v.s)^(e-v.s));
            if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
            return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
                   (d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
                   (d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
                   (d4==0 && sgn((e-v.s)*(e-v.e))<=0);
        }
        //`直线和线段相交判断`
        //`-*this line   -v seg`
        //`2 规范相交`
        //`1 非规范相交`
        //`0 不相交`
        int linecrossseg(Line v){
            int d1 = sgn((e-s)^(v.s-s));
            int d2 = sgn((e-s)^(v.e-s));
            if((d1^d2)==-2) return 2;
            return (d1==0||d2==0);
        }
        //`两直线关系`
        //`0 平行`
        //`1 重合`
        //`2 相交`
        int linecrossline(Line v){
            if((*this).parallel(v))
                return v.relation(s)==3;
            return 2;
        }
        //`求两直线的交点`
        //`要保证两直线不平行或重合`
        Point crosspoint(Line v){
            double a1 = (v.e-v.s)^(s-v.s);
            double a2 = (v.e-v.s)^(e-v.s);
            return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
        }
        //点到直线的距离
        double dispointtoline(Point p){
            return fabs((p-s)^(e-s))/length();
        }
        //点到线段的距离
        double dispointtoseg(Point p){
            if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
                return min(p.distance(s),p.distance(e));
            return dispointtoline(p);
        }
        //`返回线段到线段的距离`
        //`前提是两线段不相交,相交距离就是0了`
        double dissegtoseg(Line v){
            return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
        }
        //`返回点p在直线上的投影`
        Point lineprog(Point p){
            return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
        }
        //`返回点p关于直线的对称点`
        Point symmetrypoint(Point p){
            Point q = lineprog(p);
            return Point(2*q.x-p.x,2*q.y-p.y);
        }
    };
    int main()
    {
        int T;
        cin >> T;
        while(T--)
        {
            Line l1,l2;
            l1.input();
            l2.input();
            int flag = l1.segcrossseg(l2);
            if(flag == 0)
            {
                cout << "0.00
    ";
                continue;
            }
    
            Point p = l1.crosspoint(l2);
            if(l1.s.y < l1.e.y)
            {
                swap(l1.s,l1.e);
            }
            if(l2.s.y < l2.e.y)
            {
                swap(l2.s,l2.e);
            }
            double y = min(l1.s.y,l2.s.y);
            if(l1.s.y > l2.s.y)
            {
                swap(l1,l2);
            }
            if(fabs((l1.s-l1.e) ^ (l2.s-l2.e)) <= eps)
            {
                cout << "0.00
    ";
                continue;
            }
            Line l4 = Line(l2.s,Point(l2.s.x,100000));
            if(l4.segcrossseg(l1))
            {
                cout << "0.00
    ";
                continue;
            }
            l4 = Line(l1.s,Point(l1.s.x,100000));
            if(l4.segcrossseg(l2))
            {
                cout << "0.00
    ";
                continue;
            }
            Line l3 = Line(l1.s,Point(10000,l1.s.y));
            Point p1 = l3.crosspoint(l2);
            double ans = fabs(p.y-y) * fabs(p1.x-l1.s.x) / 2;
            printf("%0.2lf
    ",ans+eps);
        }
    
        return 0;
    }
    /*
    1
    3370 -428 -4763 -6936 -2574 -9788 5373 6696
     */
    
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  • 原文地址:https://www.cnblogs.com/hh13579/p/13859367.html
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