• poj1584(判断凸包+求点到线段的距离)


    题目链接:https://vjudge.net/problem/POJ-1584

    题意:首先要判断凸包,然后判断圆是否在多边形中。

    思路:
      判断凸包利用叉积,判断圆在多边形首先要判断圆心是否在多边形中,然后判断圆心到每条边的距离是否小于半径。板子很重要!!

    AC code:

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<cstdlib>
    using namespace std;
    
    const double eps=1e-8;
    const double inf=1e20;int sgn(double x){
        if(abs(x)<eps) return 0;
        if(x<0) return -1;
        return 1;
    }
    
    struct Point{
        double x,y;
        Point(){}
        Point(double xx,double yy):x(xx),y(yy){}
        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);
        }
        double operator * (const Point& b)const{
            return x*b.x+y*b.y;
        }
        double operator ^ (const Point& b)const{
            return x*b.y-b.x*y;
        }
        //绕原点旋转角度b(弧度值),后x、y的变化
        void transXY(double b){
            double tx=x,ty=y;
            x=tx*cos(b)-ty*sin(b);
            y=tx*sin(b)+ty*cos(b);
        }
    };
    
    struct Line{
        Point s,e;
        Line(){}
        Line(Point ss,Point ee){
            s=ss,e=ee;
        }
        //两直线相交求交点
        //第一个值为0表示直线重合,为1表示平行,为2表示相交
        //只有第一个值为2时,交点才有意义
        pair<int,Point> operator &(const Line &b)const{
            Point res = s;
            if(sgn((s-e)^(b.s-b.e)) == 0)
            {
                if(sgn((s-b.e)^(b.s-b.e)) == 0)
                    return make_pair(0,res);//重合
                else return make_pair(1,res);//平行
            }
            double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
            res.x += (e.x-s.x)*t;
            res.y += (e.y-s.y)*t;
            return make_pair(2,res);
        }
    };
    //判断线段相交
    bool inter(Line l1,Line l2){
        return
            max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)&&
            max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)&&
            max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)&&
            max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)&&
            sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0&&
            sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=0;
    }
    
    double dis(Point a,Point b){
        return sqrt((b-a)*(b-a));
    }
    //判断点在线段上
    bool OnSeg(Point P,Line L){
        return
            sgn((L.s-P)^(L.e-P))==0&&
            sgn((P.x-L.s.x)*(P.x-L.e.x))<=0&&
            sgn((P.y-L.s.y)*(P.y-L.e.y))<=0;
    }
    //判断点在凸多边形内,复杂度O(n)
    //点形成一个凸包,而且按逆时针排序(如果是顺时针把里面的<0改为>0)
    //点的编号:0~n-1
    //返回值:
    //-1:点在凸多边形外
    //0:点在凸多边形边界上
    //1:点在凸多边形内
    int inConvexPoly(Point a,Point p[],int n){
        for(int i=0;i<n;++i)
            if(sgn((p[i]-a)^(p[(i+1)%n]-a))<0) return -1;
            else if(OnSeg(a,Line(p[i],p[(i+1)%n]))) return 0;
        return 1;
    }
    //判断点在任意多边形内,复杂度O(n)
    //射线法,poly[]的顶点数要大于等于3,点的编号0~n-1,按逆时针或顺时针排序
    //返回值
    //-1:点在凸多边形外
    //0:点在凸多边形边界上
    //1:点在凸多边形内
    int inPoly(Point a,Point p[],int n){
        int cnt=0;
        Line ray,side;
        ray.s=a;
        ray.e.y=a.y;
        ray.e.x=-inf;
        for(int i=0;i<n;++i){
            side.s=p[i];
            side.e=p[(i+1)%n];
            if(OnSeg(a,side)) return 0;
            if(sgn(side.s.y-side.e.y)==0) continue;
            if(OnSeg(side.s,ray)){
                if(sgn(side.s.y-side.e.y)>0) ++cnt;
            }
            else if(OnSeg(side.e,ray)){
                if(sgn(side.e.y-side.s.y)>0) ++cnt;
            }
            else if(inter(ray,side)) ++cnt;
        }
        if(cnt%2==1) return 1;
        else return -1;
    }
    //判断凸多边形
    //允许共线边
    //点可以是顺时针给出也可以是逆时针给出
    //点的编号是0~n-1
    bool isconvex(Point poly[],int n){
        bool s[3];
        memset(s,false,sizeof(s));
        for(int i=0;i<n;++i){
            s[sgn((poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]))+1]=true;
            if(s[0]&&s[2]) return false;
        }
        return true;
    }
    //点到线段的距离
    //返回点到线段最近的点
    Point NearestPointToLineSeg(Point P,Line L){
        Point result;
        double t=((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
        if(t>=0&&t<=1){
            result.x=L.s.x+(L.e.x-L.s.x)*t;
            result.y=L.s.y+(L.e.y-L.s.y)*t;
        }
        else{
            if(dis(P,L.s)<dis(P,L.e))
                return L.s;
            else 
                return L.e;
        }
        return result;
    }
    
    int n;
    double R,X,Y;
    Point pt[110];
    
    int main(){
        while(scanf("%d",&n),n>=3){
            scanf("%lf%lf%lf",&R,&X,&Y);
            for(int i=0;i<n;++i)
                scanf("%lf%lf",&pt[i].x,&pt[i].y);
            if(!isconvex(pt,n)){
                printf("HOLE IS ILL-FORMED
    ");
                continue;
            }
            Point P=Point(X,Y);
            if(inPoly(P,pt,n)==-1){
                printf("PEG WILL NOT FIT
    ");
                continue;
            }
            int flag=1;
            for(int i=0;i<n;++i){
                if(sgn(dis(P,NearestPointToLineSeg(P,Line(pt[i],pt[(i+1)%n])))-R)<0){
                    flag=0;
                    break;
                }
            }
            if(flag) printf("PEG WILL FIT
    ");
            else printf("PEG WILL NOT FIT
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/FrankChen831X/p/11823406.html
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