计算几何模板

我们正在使用的从网上抄下来的计算几何的模板

#include <bits/stdc++.h>
using namespace std;

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;}
 
struct Point{
    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\n",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);
    }
};
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);
    }
};
//圆
struct circle{
    Point p;//圆心
    double r;//半径
circle(){}
    circle(Point _p,double _r){
        p = _p;
        r = _r;
    }
    circle(double x,double y,double _r){
        p = Point(x,y);
        r = _r;
    }
    //三角形的外接圆
    //需要 Point 的 + / rotate() 以及 Line 的 crosspoint()
    //利用两条边的中垂线得到圆心
    //测试:UVA12304
    circle(Point a,Point b,Point c){
        Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
        Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
        p = u.crosspoint(v);
        r = p.distance(a);
    }
    //三角形的内切圆
    //参数 bool t 没有作用,只是为了和上面外接圆函数区别
    //测试:UVA12304
    circle(Point a,Point b,Point c,bool t){
        Line u,v;
        double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.
        x);
        u.s = a;
        u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
        v.s = b;
        m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
        v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
        p = u.crosspoint(v);
        r = Line(a,b).dispointtoseg(p);
    }
    //输入
    void input(){
        p.input();
        scanf("%lf",&r);
    }
    //输出
    void output(){
        printf("%.2lf-%.2lf-%.2lf\n",p.x,p.y,r);
    }
    bool operator == (circle v){
        return (p==v.p) && sgn(r-v.r)==0;
    }
    bool operator < (circle v)const{
        return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
    }
    //面积
    double area(){
        return pi*r*r;
    }
    //周长
    double circumference(){
        return 2*pi*r;
    }
    //点和圆的关系
    //0 圆外
    //1 圆上
    //2 圆内
    int relation(Point b){
        double dst = b.distance(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r)==0)return 1;
        return 0;
    }
    //线段和圆的关系
    //比较的是圆心到线段的距离和半径的关系
    int relationseg(Line v){
        double dst = v.dispointtoseg(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r) == 0)return 1;
        return 0;
    }
    //直线和圆的关系
    //比较的是圆心到直线的距离和半径的关系
    int relationline(Line v){
        double dst = v.dispointtoline(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r) == 0)return 1;
        return 0;
    }
    //两圆的关系
    //5 相离
    //4 外切
    //3 相交
    //2 内切
    //1 内含
    //需要 Point 的 distance
    //测试:UVA12304
    int relationcircle(circle v){
        double d = p.distance(v.p);
        if(sgn(d-r-v.r) > 0)return 5;
        if(sgn(d-r-v.r) == 0)return 4;
        double l = fabs(r-v.r);
        if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
        if(sgn(d-l)==0)return 2;
        if(sgn(d-l)<0)return 1;
    }
    //求两个圆的交点,返回 0 表示没有交点,返回 1 是一个交点,2 是两个交点
    //需要 relationcircle
    //测试:UVA12304
    int pointcrosscircle(circle v,Point &p1,Point &p2){
        int rel = relationcircle(v);
        if(rel == 1 || rel == 5)return 0;
        double d = p.distance(v.p);
        double l = (d*d+r*r-v.r*v.r)/(2*d);
        double h = sqrt(r*r-l*l);
        Point tmp = p + (v.p-p).trunc(l);
        p1 = tmp + ((v.p-p).rotleft().trunc(h));
        p2 = tmp + ((v.p-p).rotright().trunc(h));
        if(rel == 2 || rel == 4)
        return 1;
        return 2;
    }
    //求直线和圆的交点,返回交点个数
    int pointcrossline(Line v,Point &p1,Point &p2){
        if(!(*this).relationline(v))return 0;
        Point a = v.lineprog(p);
        double d = v.dispointtoline(p);
        d = sqrt(r*r-d*d);
        if(sgn(d) == 0){
            p1 = a;
            p2 = a;
            return 1;
        }
        p1 = a + (v.e-v.s).trunc(d);
        p2 = a-(v.e-v.s).trunc(d);
        return 2;
    }
    //得到过 a,b 两点,半径为 r1 的两个圆
    int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){
        circle x(a,r1),y(b,r1);
        int t = x.pointcrosscircle(y,c1.p,c2.p);
        if(!t)return 0;
        c1.r = c2.r = r;
        return t;
    }
    //得到与直线 u 相切,过点 q, 半径为 r1 的圆
    //测试:UVA12304
    int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){
        double dis = u.dispointtoline(q);
        if(sgn(dis-r1*2)>0)return 0;
        if(sgn(dis) == 0){
            c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
            c2.p = q + ((u.e-u.s).rotright().trunc(r1));
            c1.r = c2.r = r1;
            return 2;
        }
 
        Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e +
        (u.e-u.s).rotleft().trunc(r1)));
        Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e
        + (u.e-u.s).rotright().trunc(r1)));
        circle cc = circle(q,r1);
        Point p1,p2;
        if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2)
        ;
        c1 = circle(p1,r1);
        if(p1 == p2){
            c2 = c1;
            return 1;
        }
        c2 = circle(p2,r1);
        return 2;
    }
    //同时与直线 u,v 相切,半径为 r1 的圆
    //测试:UVA12304
    int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,
    circle &c3,circle &c4){
        if(u.parallel(v))return 0;//两直线平行
        Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u
        .e-u.s).rotleft().trunc(r1));
        Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (
        u.e-u.s).rotright().trunc(r1));
        Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v
        .e-v.s).rotleft().trunc(r1));
        Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (
        v.e-v.s).rotright().trunc(r1));
        c1.r = c2.r = c3.r = c4.r = r1;
        c1.p = u1.crosspoint(v1);
        c2.p = u1.crosspoint(v2);
        c3.p = u2.crosspoint(v1);
        c4.p = u2.crosspoint(v2);
        return 4;
    }
    //同时与不相交圆 cx,cy 相切,半径为 r1 的圆
    //测试:UVA12304
    int getcircle(circle cx,circle cy,double r1,circle &c1,circle &
    c2){
        circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
        int t = x.pointcrosscircle(y,c1.p,c2.p);
        if(!t)return 0;
        c1.r = c2.r = r1;
        return t;
    }
 
    //过一点作圆的切线 (先判断点和圆的关系)
    //测试:UVA12304
    int tangentline(Point q,Line &u,Line &v){
        int x = relation(q);
        if(x == 2)return 0;
        if(x == 1){
            u = Line(q,q + (q-p).rotleft());
            v = u;
            return 1;
        }
        double d = p.distance(q);
        double l = r*r/d;
        double h = sqrt(r*r-l*l);
        u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)))
        ;
        v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h))
        );
        return 2;
    }
    //求两圆相交的面积
    double areacircle(circle v){
        int rel = relationcircle(v);
        if(rel >= 4)return 0.0;
        if(rel <= 2)return min(area(),v.area());
        double d = p.distance(v.p);
        double hf = (r+v.r+d)/2.0;
        double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
        double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
        a1 = a1*r*r;
        double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
        a2 = a2*v.r*v.r;
        return a1+a2-ss;
    }
    //求圆和三角形 pab 的相交面积
    //测试:POJ3675 HDU3982 HDU2892
    double areatriangle(Point a,Point b){
        if(sgn((p-a)^(p-b)) == 0)return 0.0;
        Point q[5];
        int len = 0;
        q[len++] = a;
        Line l(a,b);
        Point p1,p2;
        if(pointcrossline(l,q[1],q[2])==2){
            if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
            if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
        }
        q[len++] = b;
        if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q
        [2]);
        double res = 0;
        for(int i = 0;i < len-1;i++){
            if(relation(q[i])==0||relation(q[i+1])==0){
                double arg = p.rad(q[i],q[i+1]);
                res += r*r*arg/2.0;
            }
            else{
                res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
            }
        }
        return res;
    }
};
 
struct polygon{
    int n;
    Point p[maxp];
    Line l[maxp];
    void input(int _n){
        n = _n;
        for(int i = 0;i < n;i++)
        p[i].input();
    }
    void add(Point q){
        p[n++] = q;
    }
    void getline(){
        for(int i = 0;i < n;i++){
            l[i] = Line(p[i],p[(i+1)%n]);
        }
    }
    struct cmp{
        Point p;
    cmp(const Point &p0){p = p0;}
        bool operator()(const Point &aa,const Point &bb){
            Point a = aa, b = bb;
            int d = sgn((a-p)^(b-p));
            if(d == 0){
                return sgn(a.distance(p)-b.distance(p)) < 0;
            }
            return d > 0;
        }
    };
    //进行极角排序
    //首先需要找到最左下角的点
    //需要重载号好 Point 的 < 操作符 (min 函数要用)
    void norm(){
        Point mi = p[0];
        for(int i = 1;i < n;i++)mi = min(mi,p[i]);
        sort(p,p+n,cmp(mi));
    }
    //得到凸包
    //得到的凸包里面的点编号是 0~n-1 的
    //两种凸包的方法
    //注意如果有影响,要特判下所有点共点,或者共线的特殊情况
    //测试 LightOJ1203 LightOJ1239
    void getconvex(polygon &convex){
        sort(p,p+n);
        convex.n = n;
        for(int i = 0;i < min(n,2);i++){
            convex.p[i] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
        --;//特判
        if(n <= 2)return;
        int &top = convex.n;
        top = 1;
        for(int i = 2;i < n;i++){
            while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-
            p[i])) <= 0)
            top--;
            convex.p[++top] = p[i];
        }
        int temp = top;
        convex.p[++top] = p[n-2];
        for(int i = n-3;i >= 0;i--){
            while(top != temp && sgn((convex.p[top]-p[i])^(convex.p
            [top-1]-p[i])) <= 0)
            top--;
            convex.p[++top] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
        --;//特判
        convex.norm();//原来得到的是顺时针的点,排序后逆时针
    }
    //得到凸包的另外一种方法
    //测试 LightOJ1203 LightOJ1239
    void Graham(polygon &convex){
        norm();
        int &top = convex.n;
        top = 0;
        if(n == 1){
            top = 1;
            convex.p[0] = p[0];
            return;
        }
        if(n == 2){
            top = 2;
            convex.p[0] = p[0];
            convex.p[1] = p[1];
            if(convex.p[0] == convex.p[1])top--;
            return;
        }
        convex.p[0] = p[0];
        convex.p[1] = p[1];
        top = 2;
        for(int i = 2;i < n;i++){
            while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])
            ^(p[i]-convex.p[top-2])) <= 0 )
            top--;
            convex.p[top++] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n
        --;//特 判
    }
    //判断是不是凸的
    bool isconvex(){
        bool s[2];
        memset(s,false,sizeof(s));
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            int k = (j+1)%n;
            s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;
            if(s[0] && s[2])return false;
        }
        return true;
    }
    //判断点和任意多边形的关系
    // 3 点上
    // 2 边上
    // 1 内部
    // 0 外部
    int relationpoint(Point q){
        for(int i = 0;i < n;i++){
            if(p[i] == q)return 3;
        }
        getline();
        for(int i = 0;i < n;i++){
            if(l[i].pointonseg(q))return 2;
        }
        int cnt = 0;
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            int k = sgn((q-p[j])^(p[i]-p[j]));
            int u = sgn(p[i].y-q.y);
            int v = sgn(p[j].y-q.y);
            if(k > 0 && u < 0 && v >= 0)cnt++;
            if(k < 0 && v < 0 && u >= 0)cnt--;
        }
        return cnt != 0;
    }
    //直线 u 切割凸多边形左侧
    //注意直线方向
    //测试:HDU3982
    void convexcut(Line u,polygon &po){
        int &top = po.n;//注意引用
        top = 0;
        for(int i = 0;i < n;i++){
            int d1 = sgn((u.e-u.s)^(p[i]-u.s));
            int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
            if(d1 >= 0)po.p[top++] = p[i];
            if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i
            +1)%n]));
        }
    }
    //得到周长
    //测试 LightOJ1239
    double getcircumference(){
        double sum = 0;
        for(int i = 0;i < n;i++){
            sum += p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }
    //得到面积
    double getarea(){
        double sum = 0;
        for(int i = 0;i < n;i++){
            sum += (p[i]^p[(i+1)%n]);
        }
        return fabs(sum)/2;
    }
    //得到方向
    // 1 表示逆时针,0 表示顺时针
    bool getdir(){
        double sum = 0;
        for(int i = 0;i < n;i++)
        sum += (p[i]^p[(i+1)%n]);
        if(sgn(sum) > 0)return 1;
        return 0;
    }
    //得到重心
    Point getbarycentre(){
        Point ret(0,0);
        double area = 0;
        for(int i = 1;i < n-1;i++){
            double tmp = (p[i]-p[0])^(p[i+1]-p[0]);
            if(sgn(tmp) == 0)continue;
            area += tmp;
            ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;
            ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;
        }
        if(sgn(area)) ret = ret/area;
        return ret;
    }
    //多边形和圆交的面积
    //测试:POJ3675 HDU3982 HDU2892
    double areacircle(circle c){
        double ans = 0;
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)
            ans += c.areatriangle(p[i],p[j]);
            else ans-= c.areatriangle(p[i],p[j]);
        }
        return fabs(ans);
    }
    //多边形和圆关系
    // 2 圆完全在多边形内
    // 1 圆在多边形里面,碰到了多边形边界
    // 0 其它
    int relationcircle(circle c){
        getline();
        int x = 2;
        if(relationpoint(c.p) != 1)return 0;//圆心不在内部
        for(int i = 0;i < n;i++){
            if(c.relationseg(l[i])==2)return 0;
            if(c.relationseg(l[i])==1)x = 1;
        }
        return x;
    }
};
//AB X AC
double cross(Point A,Point B,Point C){
    return (B-A)^(C-A);
}
//AB*AC
double dot(Point A,Point B,Point C){
    return (B-A)*(C-A);
}
//最小矩形面积覆盖
// A 必须是凸包 (而且是逆时针顺序)
// 测试 UVA 10173
double minRectangleCover(polygon A){
    //要特判 A.n < 3 的情况
    if(A.n < 3)return 0.0;
    A.p[A.n] = A.p[0];
    double ans =-1;
    int r = 1, p = 1, q;
    for(int i = 0;i < A.n;i++){
        //卡出离边 A.p[i] - A.p[i+1] 最远的点
        while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1])-cross(A.p[i],
        A.p[i+1],A.p[r]) ) >= 0 )
        r = (r+1)%A.n;
        //卡出 A.p[i] - A.p[i+1] 方向上正向 n 最远的点
        while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1])-dot(A.p[i],A.p[i
        +1],A.p[p]) ) >= 0 )
        p = (p+1)%A.n;
        if(i == 0)q = p;
        //卡出 A.p[i] - A.p[i+1] 方向上负向最远的点
        while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1])-dot(A.p[i],A.p[i
        +1],A.p[q])) <= 0)
        q = (q+1)%A.n;
        double d = (A.p[i]-A.p[i+1]).len2();
        double tmp = cross(A.p[i],A.p[i+1],A.p[r]) *
        (dot(A.p[i],A.p[i+1],A.p[p])-dot(A.p[i],A.p[i+1],A.p[
        q]))/d;
        if(ans < 0 || ans > tmp)ans = tmp;
    }
    return ans;
}
 
 
 
//直线切凸多边形
//多边形是逆时针的,在 q1q2 的左侧
//测试:HDU3982
vector<Point> convexCut(const vector<Point> &ps,Point q1,Point q2){
    vector<Point>qs;
    int n = ps.size();
    for(int i = 0;i < n;i++){
        Point p1 = ps[i], p2 = ps[(i+1)%n];
        int d1 = sgn((q2-q1)^(p1-q1)), d2 = sgn((q2-q1)^(p2-q1));
        if(d1 >= 0)
        qs.push_back(p1);
        if(d1 * d2 < 0)
        qs.push_back(Line(p1,p2).crosspoint(Line(q1,q2)));
    }
    return qs;
}
//半平面交
//测试 POJ3335 POJ1474 POJ1279
//***************************
struct halfplane:public Line{
    double angle;
halfplane(){}
    //表示向量 s->e 逆时针 (左侧) 的半平面
    halfplane(Point _s,Point _e){
        s = _s;
        e = _e;
    }
    halfplane(Line v){
        s = v.s;
        e = v.e;
    }
    void calcangle(){
        angle = atan2(e.y-s.y,e.x-s.x);
    }
    bool operator <(const halfplane &b)const{
        return angle < b.angle;
    }
};
struct halfplanes{
    int n;
    halfplane hp[maxp];
    Point p[maxp];
    int que[maxp];
    int st,ed;
    void push(halfplane tmp){
        hp[n++] = tmp;
    }
    //去重
    void unique(){
        int m = 1;
        for(int i = 1;i < n;i++){
            if(sgn(hp[i].angle-hp[i-1].angle) != 0)
            hp[m++] = hp[i];
            else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s)
            ) > 0)
            hp[m-1] = hp[i];
        }
        n = m;
    }
    bool halfplaneinsert(){
        for(int 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(int i = 2;i < n;i++){
            while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))
            <0)ed--;
            while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))
            <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 && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[
        que[st]].s))<0)ed--;
        while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-
        hp[que[ed]].s))<0)st++;
        if(st+1>=ed)return false;
        return true;
    }
    //得到最后半平面交得到的凸多边形
    //需要先调用 halfplaneinsert() 且返回 true
    void getconvex(polygon &con){
        p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
        con.n = ed-st+1;
        for(int j = st,i = 0;j <= ed;i++,j++)
        con.p[i] = p[j];
    }
};
//***************************
 
const int maxn = 1010;
struct circles{
    circle c[maxn];
    double ans[maxn];//ans[i] 表示被覆盖了 i 次的面积
    double pre[maxn];
    int n;
    circles(){}
    void add(circle cc){
        c[n++] = cc;
    }
    //x 包含在 y 中
    bool inner(circle x,circle y){
        if(x.relationcircle(y) != 1)return 0;
        return sgn(x.r-y.r)<=0?1:0;
    }
    //圆的面积并去掉内含的圆
    void init_or(){
    bool mark[maxn] = {0};
        int i,j,k=0;
        for(i = 0;i < n;i++){
            for(j = 0;j < n;j++)
            if(i != j && !mark[j]){
                if( (c[i]==c[j])||inner(c[i],c[j]) )break;
            }
            if(j < n)mark[i] = 1;
        }
        for(i = 0;i < n;i++)
        if(!mark[i])
        c[k++] = c[i];
        n = k;
    }
    //圆的面积交去掉内含的圆
    void init_add(){
        int i,j,k;
    bool mark[maxn] = {0};
        for(i = 0;i < n;i++){
            for(j = 0;j < n;j++)
            if(i != j && !mark[j]){
                if( (c[i]==c[j])||inner(c[j],c[i]) )break;
            }
            if(j < n)mark[i] = 1;
        }
        for(i = 0;i < n;i++)
        if(!mark[i])
        c[k++] = c[i];
        n = k;
    }
    //半径为 r 的圆,弧度为 th 对应的弓形的面积
    double areaarc(double th,double r){
        return 0.5*r*r*(th-sin(th));
    }
    //测试 SPOJVCIRCLES SPOJCIRUT
    //SPOJVCIRCLES 求 n 个圆并的面积,需要加上 init_or() 去掉重复圆(否则WA)
    //SPOJCIRUT 是求被覆盖 k 次的面积,不能加 init_or()
    //对于求覆盖多少次面积的问题,不能解决相同圆,而且不能 init_or()
    //求多圆面积并,需要 init_or, 其中一个目的就是去掉相同圆
    void getarea(){
        memset(ans,0,sizeof(ans));
        vector<pair<double,int> >v;
        for(int i = 0;i < n;i++){
            v.clear();
            v.push_back(make_pair(-pi,1));
            v.push_back(make_pair(pi,-1));
            for(int j = 0;j < n;j++)
            if(i != j){
                Point q = (c[j].p-c[i].p);
                double ab = q.len(),ac = c[i].r, bc = c[j].r;
                if(sgn(ab+ac-bc)<=0){
                    v.push_back(make_pair(-pi,1));
                    v.push_back(make_pair(pi,-1));
                    continue;
                }
                if(sgn(ab+bc-ac)<=0)continue;
                if(sgn(ab-ac-bc)>0)continue;
                double th = atan2(q.y,q.x), fai = acos((ac*ac+
                ab*ab-bc*bc)/(2.0*ac*ab));
                double a0 = th-fai;
                if(sgn(a0+pi)<0)a0+=2*pi;
                double a1 = th+fai;
                if(sgn(a1-pi)>0)a1-=2*pi;
                if(sgn(a0-a1)>0){
                    v.push_back(make_pair(a0,1));
                    v.push_back(make_pair(pi,-1));
                    v.push_back(make_pair(-pi,1));
                    v.push_back(make_pair(a1,-1));
                }
                else{
                    v.push_back(make_pair(a0,1));
                    v.push_back(make_pair(a1,-1));
                }
            }
            sort(v.begin(),v.end());
            int cur = 0;
            for(int j = 0;j < v.size();j++){
                if(cur && sgn(v[j].first-pre[cur])){
                    ans[cur] += areaarc(v[j].first-pre[cur],c[i].r)
                    ;
                    ans[cur] += 0.5*(Point(c[i].p.x+c[i].r*cos(pre[
                    cur]),c[i].p.y+c[i].r*sin(pre[cur]))^Point(c
                    [i].p.x+c[i].r*cos(v[j].first),c[i].p.y+c[i
                    ].r*sin(v[j].first)));
                }
                cur += v[j].second;
                pre[cur] = v[j].first;
            }
        }
        for(int i = 1;i < n;i++)
        ans[i]-= ans[i+1];
    }
};
 
//.3平面最近点对
const int MAXN = 100010;
const double INF = 1e20;
//struct Point{
//    double x,y;
//    void input(){
//        scanf("%lf%lf",&x,&y);
//    }
//};
double dist(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
Point p[MAXN];
Point tmpt[MAXN];
bool cmpx(Point a,Point b){
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool cmpy(Point a,Point b){
    return a.y < b.y || (a.y == b.y && a.x < b.x);
}
double Closest_Pair(int left,int right){
    double d = INF;
    if(left == right)return d;
    if(left+1 == right)return dist(p[left],p[right]);
    int mid = (left+right)/2;
    double d1 = Closest_Pair(left,mid);
    double d2 = Closest_Pair(mid+1,right);
    d = min(d1,d2);
    int cnt = 0;
    for(int i = left;i <= right;i++){
        if(fabs(p[mid].x-p[i].x) <= d)
        tmpt[cnt++] = p[i];
    }
    sort(tmpt,tmpt+cnt,cmpy);
    for(int i = 0;i < cnt;i++){
        for(int j = i+1;j < cnt && tmpt[j].y-tmpt[i].y < d;j++)
        d = min(d,dist(tmpt[i],tmpt[j]));
    }
    return d;
}

int main()
{
    
    return 0; 
 } 
Last modification:February 15th, 2019 at 02:30 pm
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