poj2318

题意：把一个矩形划分成n部分，再给出一堆点，求每个部分内落入了多少点

 

sol attempt1：暴力

注意每个polygon中的点要按笔画的顺序排列好。还有就是有的点可能落在了upper or lower edge，稍微处理一下（ans==1 or 2都算）

TLE了 -_-||

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<iostream>
#include<sstream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<climits>
#include<complex>
#define mp make_pair
#define pb push_back
#define MAXN 5010
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
const double inf=1e20;
const int maxp=10;   //多边形内点的数量，注意按需求调整

int sgn(double x)
{
    if (fabs(x)<eps)    return 0;
    if (x<0)    return -1;
        else return 1;
}

int dblcmp(double d)
{
    if (fabs(d)<eps)return 0;
    return d>eps?1:-1;
}

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 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;
    }

    point operator +(const point &b)const
    {
        return point(x+b.x,y+b.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);
    }
    double operator ^(const point &b)const
    {
        return x*b.y-y*b.x;
    }
    double operator *(const point &b)const
    {
        return x*b.x+y*b.y;
    }

    double len()
    {
        return hypot(x,y);
    }
    double len2()
    {
        return x*x+y*y;
    }
    double distance(point p)
    {
        return hypot(x-p.x,y-p.y);
    }
    point add(point p)
    {
        return point(x+p.x,y+p.y);
    }
    point sub(point p)
    {
        return point(x-p.x,y-p.y);
    }
    point mul(double b)
    {
        return point(x*b,y*b);
    }
    point div(double b)
    {
        return point(x/b,y/b);
    }
    double dot(point p)
    {
        return x*p.x+y*p.y;
    }
    double det(point p)
    {
        return x*p.y-y*p.x;
    }
    double rad(point a,point b)
    {
        point p=*this;
        return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
    }
    point trunc(double r)
    {
        double l=len();
        if (!dblcmp(l))return *this;
        r/=l;
        return point(x*r,y*r);
    }
    point rotleft()
    {
        return point(-y,x);
    }
    point rotright()
    {
        return point(y,-x);
    }
    point rotate(point p,double angle)//绕点p逆时针旋转angle角度
    {
        point v=this->sub(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 a,b;
    line(){}
    line(point _a,point _b)
    {
        a=_a;
        b=_b;
    }
    bool operator==(line v)
    {
        return (a==v.a)&&(b==v.b);
    }
    //倾斜角angle
    line(point p,double angle)
    {
        a=p;
        if (dblcmp(angle-pi/2)==0)
        {
            b=a.add(point(0,1));
        }
        else
        {
            b=a.add(point(1,tan(angle)));
        }
    }
    //ax+by+c=0
    line(double _a,double _b,double _c)
    {
        if (dblcmp(_a)==0)
        {
            a=point(0,-_c/_b);
            b=point(1,-_c/_b);
        }
        else if (dblcmp(_b)==0)
        {
            a=point(-_c/_a,0);
            b=point(-_c/_a,1);
        }
        else
        {
            a=point(0,-_c/_b);
            b=point(1,(-_c-_a)/_b);
        }
    }
    void input()
    {
        a.input();
        b.input();
    }
    void adjust()
    {
        if (b<a)swap(a,b);
    }
    double length()
    {
        return a.distance(b);
    }
    double angle()//直线倾斜角 0<=angle<180
    {
        double k=atan2(b.y-a.y,b.x-a.x);
        if (dblcmp(k)<0)k+=pi;
        if (dblcmp(k-pi)==0)k-=pi;
        return k;
    }
    //点和线段关系
    //1 在逆时针
    //2 在顺时针
    //3 平行
    int relation(point p)
    {
        int c=dblcmp(p.sub(a).det(b.sub(a)));
        if (c<0)return 1;
        if (c>0)return 2;
        return 3;
    }
    bool pointonseg(point p)
    {
        return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
    }
    bool parallel(line v)
    {
        return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
    }
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(line v)
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
        int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
        if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
        return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
                d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
                d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
                d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
    }
    int linecrossseg(line v)//*this seg v line
    {
        int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
        int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
        if ((d1^d2)==-2)return 2;
        return (d1==0||d2==0);
    }
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(line v)
    {
        if ((*this).parallel(v))
        {
            return v.relation(a)==3;
        }
        return 2;
    }
    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));
    }
    double dispointtoline(point p)
    {
        return fabs(p.sub(a).det(b.sub(a)))/length();
    }
    double dispointtoseg(point p)
    {
        if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
        {
            return min(p.distance(a),p.distance(b));
        }
        return dispointtoline(p);
    }
    point lineprog(point p)
    {
        return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a))/b.sub(a).len2()));
    }
    point symmetrypoint(point p)
    {
        point q=lineprog(p);
        return point(2*q.x-p.x,2*q.y-p.y);
    }
};

struct Vector:public point
{
    Vector(){}
    Vector(double a,double b)
    {
        x=a;    y=b;
    }
    Vector(point _a,point _b)   //a->b
    {
        double dx=_b.x-_a.x;
        double dy=_b.y-_a.y;
        x=dx;   y=dy;
    }
    Vector(line v)
    {
        double dx=v.b.x-v.a.x;
        double dy=v.b.y-v.a.y;
        x=dx;   y=dy;
    }
    double length()
    {
        return (sqrt(x*x+y*y));
    }
    Vector Normal()
    {
        double L=sqrt(x*x+y*y);
        Vector Vans=Vector(-y/L,x/L);
        return Vans;
    }
};

struct circle
{
    point p;
    double r;
    circle(){}
    circle(point _p,double _r):
    p(_p),r(_r){};
    circle(double x,double y,double _r):
    p(point(x,y)),r(_r){};
    circle(point a,point b,point c)//三角形的外接圆
    {
        p=line(a.add(b).div(2),a.add(b).div(2).add(b.sub(a).rotleft())).crosspoint(line(c.add(b).div(2),c.add(b).div(2).add(b.sub(c).rotleft())));
        r=p.distance(a);
    }
    circle(point a,point b,point c,bool t)//三角形的内切圆
    {
        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.a=a;
        u.b=u.a.add(point(cos((n+m)/2),sin((n+m)/2)));
        v.a=b;
        m=atan2(a.y-b.y,a.x-b.x),n=atan2(c.y-b.y,c.x-b.x);
        v.b=v.a.add(point(cos((n+m)/2),sin((n+m)/2)));
        p=u.crosspoint(v);
        r=line(a,b).dispointtoseg(p);
    }
    void input()
    {
        p.input();
        scanf("%lf",&r);
    }
    void output()
    {
        printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);
    }
    bool operator==(circle v)
    {
        return ((p==v.p)&&dblcmp(r-v.r)==0);
    }
    bool operator<(circle v)const
    {
        return ((p<v.p)||(p==v.p)&&dblcmp(r-v.r)<0);
    }
    double area()
    {
        return pi*sqr(r);
    }
    double circumference()
    {
        return 2*pi*r;
    }
    //0 圆外
    //1 圆上
    //2 圆内
    int relation(point b)
    {
        double dst=b.distance(p);
        if (dblcmp(dst-r)<0)return 2;
        if (dblcmp(dst-r)==0)return 1;
        return 0;
    }
    int relationseg(line v)
    {
        double dst=v.dispointtoseg(p);
        if (dblcmp(dst-r)<0)return 2;
        if (dblcmp(dst-r)==0)return 1;
        return 0;
    }
    int relationline(line v)
    {
        double dst=v.dispointtoline(p);
        if (dblcmp(dst-r)<0)return 2;
        if (dblcmp(dst-r)==0)return 1;
        return 0;
    }
    //过a b两点 半径r的两个圆
    int getcircle(point a,point b,double r,circle&c1,circle&c2)
    {
        circle x(a,r),y(b,r);
        int t=x.pointcrosscircle(y,c1.p,c2.p);
        if (!t)return 0;
        c1.r=c2.r=r;
        return t;
    }
    //与直线u相切 过点q 半径r1的圆
    int getcircle(line u,point q,double r1,circle &c1,circle &c2)
    {
        double dis=u.dispointtoline(q);
        if (dblcmp(dis-r1*2)>0)return 0;
        if (dblcmp(dis)==0)
        {
            c1.p=q.add(u.b.sub(u.a).rotleft().trunc(r1));
            c2.p=q.add(u.b.sub(u.a).rotright().trunc(r1));
            c1.r=c2.r=r1;
            return 2;
        }
        line u1=line(u.a.add(u.b.sub(u.a).rotleft().trunc(r1)),u.b.add(u.b.sub(u.a).rotleft().trunc(r1)));
        line u2=line(u.a.add(u.b.sub(u.a).rotright().trunc(r1)),u.b.add(u.b.sub(u.a).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r1的圆
    int getcircle(line u,line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4)
    {
        if (u.parallel(v))return 0;
        line u1=line(u.a.add(u.b.sub(u.a).rotleft().trunc(r1)),u.b.add(u.b.sub(u.a).rotleft().trunc(r1)));
        line u2=line(u.a.add(u.b.sub(u.a).rotright().trunc(r1)),u.b.add(u.b.sub(u.a).rotright().trunc(r1)));
        line v1=line(v.a.add(v.b.sub(v.a).rotleft().trunc(r1)),v.b.add(v.b.sub(v.a).rotleft().trunc(r1)));
        line v2=line(v.a.add(v.b.sub(v.a).rotright().trunc(r1)),v.b.add(v.b.sub(v.a).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
    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;
    }
    int pointcrossline(line v,point &p1,point &p2)//求与线段交要先判断relationseg
    {
        if (!(*this).relationline(v))return 0;
        point a=v.lineprog(p);
        double d=v.dispointtoline(p);
        d=sqrt(r*r-d*d);
        if (dblcmp(d)==0)
        {
            p1=a;
            p2=a;
            return 1;
        }
        p1=a.sub(v.b.sub(v.a).trunc(d));
        p2=a.add(v.b.sub(v.a).trunc(d));
        return 2;
    }
    //5 相离
    //4 外切
    //3 相交
    //2 内切
    //1 内含
    int relationcircle(circle v)
    {
        double d=p.distance(v.p);
        if (dblcmp(d-r-v.r)>0)return 5;
        if (dblcmp(d-r-v.r)==0)return 4;
        double l=fabs(r-v.r);
        if (dblcmp(d-r-v.r)<0&&dblcmp(d-l)>0)return 3;
        if (dblcmp(d-l)==0)return 2;
        if (dblcmp(d-l)<0)return 1;
    }
    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+(sqr(r)-sqr(v.r))/d)/2;
        double h=sqrt(sqr(r)-sqr(l));
        p1=p.add(v.p.sub(p).trunc(l).add(v.p.sub(p).rotleft().trunc(h)));
        p2=p.add(v.p.sub(p).trunc(l).add(v.p.sub(p).rotright().trunc(h)));
        if (rel==2||rel==4)
        {
            return 1;
        }
        return 2;
    }
    //过一点做圆的切线 (先判断点和圆关系)
    int tangentline(point q,line &u,line &v)
    {
        int x=relation(q);
        if (x==2)return 0;
        if (x==1)
        {
            u=line(q,q.add(q.sub(p).rotleft()));
            v=u;
            return 1;
        }
        double d=p.distance(q);
        double l=sqr(r)/d;
        double h=sqrt(sqr(r)-sqr(l));
        u=line(q,p.add(q.sub(p).trunc(l).add(q.sub(p).rotleft().trunc(h))));
        v=line(q,p.add(q.sub(p).trunc(l).add(q.sub(p).rotright().trunc(h))));
        return 2;
    }
    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;
    }
    double areatriangle(point a,point b)
    {
        if (dblcmp(p.sub(a).det(p.sub(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 (dblcmp(a.sub(q[1]).dot(b.sub(q[1])))<0)q[len++]=q[1];
            if (dblcmp(a.sub(q[2]).dot(b.sub(q[2])))<0)q[len++]=q[2];
        }
        q[len++]=b;
        if (len==4&&(dblcmp(q[0].sub(q[1]).dot(q[2].sub(q[1])))>0))swap(q[1],q[2]);
        double res=0;
        int i;
        for (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].sub(p).det(q[i+1].sub(p))/2.0);
            }
        }
        return res;
    }
};

struct polygon
{
    int n;
    point p[maxp];
    line l[maxp];
    void input(int X)
    {
        n=X;
        for (int i=0;i<n;i++)
        {
            p[i].input();
        }
    }
    void add(point q)
    {
        p[n++]=q;
    }
    //注意：点必须按顺序读入，否则在getline的时候会混乱掉
    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=dblcmp(a.sub(p).det(b.sub(p)));
            if (d==0)
            {
                return dblcmp(a.distance(p)-b.distance(p))<0;
            }
            return d>0;
        }
    };
    void norm()
    {
        point mi=p[0];
        for (int i=1;i<n;i++)mi=min(mi,p[i]);
        sort(p,p+n,cmp(mi));
    }
    void getconvex(polygon &convex)
    {
        int i,j,k;
        sort(p,p+n);
        convex.n=n;
        for (i=0;i<min(n,2);i++)
        {
            convex.p[i]=p[i];
        }
        if (n<=2)return;
        int &top=convex.n;
        top=1;
        for (i=2;i<n;i++)
        {
            while (top&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
        int temp=top;
        convex.p[++top]=p[n-2];
        for (i=n-3;i>=0;i--)
        {
            while (top!=temp&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
    }
    bool isconvex()
    {
        bool s[3];
        memset(s,0,sizeof(s));
        int i,j,k;
        for (i=0;i<n;i++)
        {
            j=(i+1)%n;
            k=(j+1)%n;
            s[dblcmp(p[j].sub(p[i]).det(p[k].sub(p[i])))+1]=1;
            if (s[0]&&s[2])return 0;
        }
        return 1;
    }
    //3 点上
    //2 边上
    //1 内部
    //0 外部
    int relationpoint(point q)
    {
        int i,j;
        for (i=0;i<n;i++)
        {
            if (p[i]==q)return 3;
        }
        getline();
        for (i=0;i<n;i++)
        {
            if (l[i].pointonseg(q))
            {
                //cout<<"222:      "<<i<<"  "<<l[i].a.x<<","<<l[i].a.y<<"---"<<l[i].b.x<<","<<l[i].b.y<<endl;
                return 2;
            }
        }
        int cnt=0;
        for (i=0;i<n;i++)
        {
            j=(i+1)%n;
            int k=dblcmp(q.sub(p[j]).det(p[i].sub(p[j])));
            int u=dblcmp(p[i].y-q.y);
            int v=dblcmp(p[j].y-q.y);
            if (k>0&&u<0&&v>=0)cnt++;
            if (k<0&&v<0&&u>=0)cnt--;
        }
        return cnt!=0;
    }
    //1 在多边形内长度为正
    //2 相交或与边平行
    //0 无任何交点
    int relationline(line u)
    {
        int i,j,k=0;
        getline();
        for (i=0;i<n;i++)
        {
            if (l[i].segcrossseg(u)==2)return 1;
            if (l[i].segcrossseg(u)==1)k=1;
        }
        if (!k)return 0;
        vector<point>vp;
        for (i=0;i<n;i++)
        {
            if (l[i].segcrossseg(u))
            {
                if (l[i].parallel(u))
                {
                    vp.pb(u.a);
                    vp.pb(u.b);
                    vp.pb(l[i].a);
                    vp.pb(l[i].b);
                    continue;
                }
                vp.pb(l[i].crosspoint(u));
            }
        }
        sort(vp.begin(),vp.end());
        int sz=vp.size();
        for (i=0;i<sz-1;i++)
        {
            point mid=vp[i].add(vp[i+1]).div(2);
            if (relationpoint(mid)==1)return 1;
        }
        return 2;
    }
    //直线u切割凸多边形左侧
    //注意直线方向
    void convexcut(line u,polygon &po)
    {
        int i,j,k;
        int &top=po.n;
        top=0;
        for (i=0;i<n;i++)
        {
            int d1=dblcmp(p[i].sub(u.a).det(u.b.sub(u.a)));
            int d2=dblcmp(p[(i+1)%n].sub(u.a).det(u.b.sub(u.a)));
            if (d1>=0)po.p[top++]=p[i];
            if (d1*d2<0)po.p[top++]=u.crosspoint(line(p[i],p[(i+1)%n]));
        }
    }
    double getcircumference()
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }
    double getarea()
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].det(p[(i+1)%n]);
        }
        return fabs(sum)/2;
    }
    bool getdir()//1代表逆时针 0代表顺时针
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].det(p[(i+1)%n]);
        }
        if (dblcmp(sum)>0)return 1;
        return 0;
    }
    point getbarycentre()
    {
        point ret(0,0);
        double area=0;
        int i;
        for (i=1;i<n-1;i++)
        {
            double tmp=p[i].sub(p[0]).det(p[i+1].sub(p[0]));
            if (dblcmp(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 (dblcmp(area))ret=ret.div(area);
        return ret;
    }
    double areaintersection(polygon po)
    {
    }
    double areaunion(polygon po)
    {
        return getarea()+po.getarea()-areaintersection(po);
    }
    double areacircle(circle c)
    {
        int i,j,k,l,m;
        double ans=0;
        for (i=0;i<n;i++)
        {
            int j=(i+1)%n;
            if (dblcmp(p[j].sub(c.p).det(p[i].sub(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
    //0 一部分在圆外
    //1 与圆某条边相切
    //2 完全在圆内
    int relationcircle(circle c)
    {
        getline();
        int i,x=2;
        if (relationpoint(c.p)!=1)return 0;
        for (i=0;i<n;i++)
        {
            if (c.relationseg(l[i])==2)return 0;
            if (c.relationseg(l[i])==1)x=1;
        }
        return x;
    }
    void find(int st,point tri[],circle &c)
    {
        if (!st)
        {
            c=circle(point(0,0),-2);
        }
        if (st==1)
        {
            c=circle(tri[0],0);
        }
        if (st==2)
        {
            c=circle(tri[0].add(tri[1]).div(2),tri[0].distance(tri[1])/2.0);
        }
        if (st==3)
        {
            c=circle(tri[0],tri[1],tri[2]);
        }
    }
    void solve(int cur,int st,point tri[],circle &c)
    {
        find(st,tri,c);
        if (st==3)return;
        int i;
        for (i=0;i<cur;i++)
        {
            if (dblcmp(p[i].distance(c.p)-c.r)>0)
            {
                tri[st]=p[i];
                solve(i,st+1,tri,c);
            }
        }
    }
    circle mincircle()//点集最小圆覆盖
    {
        random_shuffle(p,p+n);
        point tri[4];
        circle c;
        solve(n,0,tri,c);
        return c;
    }
    int circlecover(double r)//单位圆覆盖
    {
        int ans=0,i,j;
        vector<pair<double,int> >v;
        for (i=0;i<n;i++)
        {
            v.clear();
            for (j=0;j<n;j++)if (i!=j)
            {
                point q=p[i].sub(p[j]);
                double d=q.len();
                if (dblcmp(d-2*r)<=0)
                {
                    double arg=atan2(q.y,q.x);
                    if (dblcmp(arg)<0)arg+=2*pi;
                    double t=acos(d/(2*r));
                    v.push_back(make_pair(arg-t+2*pi,-1));
                    v.push_back(make_pair(arg+t+2*pi,1));
                }
            }
            sort(v.begin(),v.end());
            int cur=0;
            for (j=0;j<v.size();j++)
            {
                if (v[j].second==-1)++cur;
                else --cur;
                ans=max(ans,cur);
            }
        }
        return ans+1;
    }
    int pointinpolygon(point q)//点在凸多边形内部的判定
    {
        if (getdir())reverse(p,p+n);
        if (dblcmp(q.sub(p[0]).det(p[n-1].sub(p[0])))==0)
        {
            if (line(p[n-1],p[0]).pointonseg(q))return n-1;
            return -1;
        }
        int low=1,high=n-2,mid;
        while (low<=high)
        {
            mid=(low+high)>>1;
            if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>=0&&dblcmp(q.sub(p[0]).det(p[mid+1].sub(p[0])))<0)
            {
                polygon c;
                c.p[0]=p[mid];
                c.p[1]=p[mid+1];
                c.p[2]=p[0];
                c.n=3;
                if (c.relationpoint(q))return mid;
                return -1;
            }
            if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>0)
            {
                low=mid+1;
            }
            else
            {
                high=mid-1;
            }
        }
        return -1;
    }
};


struct polygon PL[MAXN];
struct point   PO[MAXN];
int num[MAXN];
int n,m;
double X1,Y1,X2,Y2;
double L[MAXN],U[MAXN];

int main()
{
    while(scanf("%d",&n))
    {
        if(n==0)    break;
        scanf("%d%lf%lf%lf%lf",&m,&X1,&Y1,&X2,&Y2);
        for (int i=0; i<n; i++)
            scanf("%lf%lf",&U[i],&L[i]);
        for (int i=0; i<m; i++)
            PO[i].input();

        sort(U,U+n);
        sort(L,L+n);

        for (int i=0; i<=n; i++)
        {
            int ya=Y1,yb=Y2,yc=Y2,yd=Y1;
            int xa,xb,xc,xd;
            if (i==0)
            {
                xa=X1,xb=X1;
                xc=L[0],xd=U[0];
            }
            else if (i==n)
            {
                xa=U[n-1],xb=L[n-1];
                xc=X2,xd=X2;
            }
            else
            {
                xa=U[i-1],xb=L[i-1];
                xc=L[i],xd=U[i];
            }
            PL[i].n=4;
            PL[i].p[0]=point(xa,ya);
            PL[i].p[1]=point(xb,yb);
            PL[i].p[2]=point(xc,yc);
            PL[i].p[3]=point(xd,yd);
            PL[i].getline();
        }
        //PL[0..n]:polygon

        memset(num,0,sizeof(num));
        for (int i=0;i<=n;i++)
        {
            polygon PY=PL[i];
            PY.getline();
/*
            cout<<"PL  ";
            for (int j=0;j<=3;j++)
                cout<<PY.p[j].x<<","<<PY.p[j].y<<" ";
            cout<<endl;
*/
            for (int j=0;j<m;j++)
            {
                point PPP=PO[j];
                int ans=PY.relationpoint(PPP);
                //cout<<PPP.x<<","<<PPP.y<<"    "<<ans<<endl;
                if ((ans==1)||(ans==2))
                    num[i]++;
            }

        }

        for (int i=0; i<=n; i++)
            printf("%d: %d\n",i,num[i]);
            //cout<<i<<": "<<num[i]<<endl;
    }
    return 0;
}

 

sol attemp2：可以二分啊

对于每一个点，判断当前线段在点的左边还是右边