POJ 2318 TOYS(点在多边形内判定 刘汝佳模板)

POJ 2318 TOYS(点在多边形内判定)

http://poj.org/problem?id=2318

题意:

       有一个平行于坐标轴的长矩形,被n块木板分成了n+1个包间.然后给你一些点的坐标,问你每个包间各包含了几个点?

分析:

       直接求出每个包间的4个点坐标(按时针顺序),然后对于每个点,用点在多边形内的模板直接判定即可.

    判断点在多边形内可以直接判断4个叉积即可,因为每个多边形都是凸4边行.

这题有直接普通的遍历判断差积的话时间1630ms，时间非常久，如果用二分去做的话只有63ms！！！

直接遍历代码：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double eps=1e-10;
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
const int maxn=5000+10;
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
};
typedef Point Vector;
Vector operator-(Point A,Point B)
{
    return Vector(A.x-B.x,A.y-B.y);
}
double Dot(Vector A,Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
bool InSegment(Point P,Point A,Point B)
{
    return dcmp(Cross(A-B,P-A))==0 && dcmp(Dot(A-P,B-P))<=0;
}
/*
//本题由于都是4边行,必须加快速度判断.
//如果直接用下面这个函数模板,将超时
bool IsPointInPolygon(Point p,Point *poly,int n)
{
    int wn=0;
    for(int i=0;i<n;++i)
    {
        if(InSegment(p, poly[(i+1)%n], poly[i]) ) return true;
        int k=dcmp( Cross(poly[(i+1)%n]-poly[i], p-poly[i] ) );
        int d1=dcmp( poly[i].y-p.y );
        int d2=dcmp( poly[(i+1)%n].y-p.y );
        if(k>0 && d1<=0 && d2>0) ++wn;
        if(k<0 && d2<=0 && d1>0) --wn;
    }
    if(wn!=0) return true;
    return false;
}
*/
bool IsPointInPolygon(Point p,Point *poly)
{
    for(int i=0;i<4;++i)
        if(dcmp (Cross(poly[(i+1)%4]-poly[i], p-poly[i]) )>0 ) return false;
    return true;
}


Point p[maxn];//需要判断位置的所有点
Point up[maxn],down[maxn];//记录上一排的n+1个点,和下一排的n+1个点
Point poly[maxn][4];
int ans[maxn];//保存第i个隔间有几个点

int main()
{
    int n,m;
    double x1,y1,x2,y2;
    bool first=true;
    while(scanf("%d",&n)==1 && n)
    {
        if(!first) printf("\n");
        first=false;

        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        up[0]=Point(x1,y1);
        up[n+1]=Point(x2,y1);
        down[0]=Point(x1,y2);
        down[n+1]=Point(x2,y2);
        for(int i=1;i<=n;++i)
        {
            scanf("%lf%lf",&x1,&x2);
            up[i]=Point(x1,y1);
            down[i]=Point(x2,y2);
        }

        for(int i=0;i<=n;++i)
        {
            poly[i][0]=up[i];
            poly[i][1]=up[i+1];
            poly[i][2]=down[i+1];
            poly[i][3]=down[i];
        }
        memset(ans,0,sizeof(ans));

        for(int i=1;i<=m;++i)
        {
            double x,y;
            scanf("%lf%lf",&x,&y);
            for(int j=0;j<=n;++j)
            if(IsPointInPolygon(Point(x,y), poly[j] ))
            {
                ans[j]++;
                break;
            }
        }
        for(int i=0;i<=n;++i)
            printf("%d: %d\n",i,ans[i]);
    }
    return 0;
}

二分查找 63ms：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double eps=1e-10;
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
const int maxn=5000+10;
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
};
typedef Point Vector;
Vector operator-(Point A,Point B)
{
    return Vector(A.x-B.x,A.y-B.y);
}
double Dot(Vector A,Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}

Point p[maxn];//需要判断位置的所有点
Point up[maxn],down[maxn];//记录上一排的n+1个点,和下一排的n+1个点
Point poly[maxn][4];
int ans[maxn];//保存第i个隔间有几个点

int main()
{
    int n,m;
    double x1,y1,x2,y2;
    bool first=true;
    while(scanf("%d",&n)==1 && n)
    {
        if(!first) printf("\n");
        first=false;

        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        up[0]=Point(x1,y1);
        up[n+1]=Point(x2,y1);
        down[0]=Point(x1,y2);
        down[n+1]=Point(x2,y2);
        for(int i=1;i<=n;++i)
        {
            scanf("%lf%lf",&x1,&x2);
            up[i]=Point(x1,y1);
            down[i]=Point(x2,y2);
        }
        memset(ans,0,sizeof(ans));

        for(int i=1;i<=m;++i)
        {
            double x,y;
            scanf("%lf%lf",&p[i].x,&p[i].y);
            int l = 1,r = 1+n;
            int tmp;
            while( l <= r)
            {
                int mid = (l + r)/2;
                if(dcmp (Cross(up[mid]-down[mid], p[i]-down[mid]) )>0)
                {
                    tmp = mid;
                    r = mid - 1;
                }
                else l = mid + 1;
            }
            ans[tmp-1]++;
        }
        for(int i=0;i<=n;++i)
            printf("%d: %d\n",i,ans[i]);
    }
    return 0;
}