ZOJ 1081 Points Within(点在多边形内判定)

ZOJ 1081 Points Within(点在多边形内判定)

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=81

题意:

       给你一个简单多边形,然后给你m个点,问你这m个点分别是否在多边形内部(包括边界上)?

分析:

       直接用刘汝佳的判定点在多边形内部的模板即可.不再赘述了.

#include<cstdio>#include<cmath>#include<cstring>using namespace std;const double eps=1e-10;const int maxn=1000+10;int dcmp(double x){    if(fabs(x)<eps) return 0;    return x<0?-1:1;}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;}bool PointInPolygon(Point p,Point* poly,int n){    int wn=0;    for(int i=0;i<n;++i)    {        if(InSegment(p,poly[i],poly[(i+1)%n])) 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;}int main(){    int n,m,kase=0;    while(scanf("%d",&n)==1 && n)    {        if(kase>0) printf("\n");        printf("Problem %d:\n",++kase);        scanf("%d",&m);        Point poly[maxn];        for(int i=0;i<n;++i)        {            scanf("%lf%lf",&poly[i].x,&poly[i].y);        }        for(int i=0;i<m;++i)        {            Point p;            scanf("%lf%lf",&p.x,&p.y);            if(PointInPolygon(p,poly,n)) printf("Within\n");            else printf("Outside\n");        }    }    return 0;}