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;}