【POJ 2318】TOYS（二分+叉积）

传送门

设点a 假如a在一个板子1的左边 那他肯定也在一个板子2左边 而且板子还是按顺序排的

满足单调性 可以二分

考虑check 判断点a在直线L左边 只需判断a到直线L的端点1 端点2两个向量的叉积小不小于0  

#include<iostream>
#include<cmath>
#include<iomanip>
#include<cstdio>
#include<cstring>
#define eps 1e-6
#define N 5005
using namespace std;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){};
}p[N];
typedef Point Vector;
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 Vector &a,const Vector &b)
{
    return a.x<b.x||(a.x==b.x)&&(a.y<b.y);
}
double Dot(Vector a,Vector b)//点积 
{
    return a.x*b.x+a.y*b.y;
}
double Len(Vector a)
{
    return sqrt(Dot(a,a));
}
double Angle(Vector a,Vector b)//夹角 
{
    return acos(Dot(a,b)/Len(a)/Len(b));
}
double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}
int dcmp(double x)
{
    if(fabs(x)<eps)  return 0;
    else return x<0?-1:1;
}
struct Line
{
    Point p1,p2;
    Line(){};
    Line(Point p1,Point p2):p1(p1),p2(p2){}
}l[N];
int n,m;
Point upper_left,lower_right;
inline bool PointIsLeft(Point p,Line k)
{
    Vector v1=k.p1-p;
    Vector v2=k.p2-p;
    return dcmp(Cross(v1,v2))<0;
}
inline int BinarySearch(Point p)
{
    int le=0,ri=n-1;
    while(le<=ri)
    {
        int m=(le+ri)>>1;
        if(PointIsLeft(p,l[m]))    ri=m-1;
        else le=m+1;
    }
    return le;
}
int ans[N];
int main()
{
    while(true)
    {
        cin>>n;
        if(!n)    break;
        cin>>m>>upper_left.x>>upper_left.y>>lower_right.x>>lower_right.y;
        double x1,x2;
        for(int i=0;i<n;i++)
        {
            cin>>x1>>x2;
            Point p1=Point(x1,upper_left.y);
            Point p2=Point(x2,lower_right.y);
            l[i]=Line(p1,p2);
        }
        memset(ans,0,sizeof(ans));
        for(int i=0;i<m;i++)
        {
            cin>>p[i].x>>p[i].y;
            int a=BinarySearch(p[i]);
            //cout<<a<<endl;
            ans[a]++; 
        }
        for(int i=0;i<n+1;i++)
        {
            cout<<i<<": "<<ans[i]<<'\n';
        }
        cout<<'\n';
    }

}