• uva10256 凸包


    链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1197

    题意:平面上n个红点,m个蓝点,问是否存在一条直线,使得任取一个红点和一个蓝点都在直线异侧。

    思路:可以把《训练指南》上的两个思路换一下顺序,求出两个凸包之后,先判断一个凸包上的点是否在另一个凸包内侧,不需要判断所有红色的点是否在蓝色凸包内,只要判断红凸包上的顶点就够了。再来判断线段相交,这时只要判断是否是规范相交就好了。觉得数据有点水。

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=510;
    const double eps=1e-8;
    struct Point
    {
        int x,y;
        Point(int x=0,int y=0):x(x),y(y) {}
    };
    int n,m;
    Point red[maxn],blue[maxn];
    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);
    }
    double Dot(Vector A,Vector B)
    {
        return A.x*B.x+A.y*B.y;
    }
    
    double Length(Vector A)
    {
        return sqrt(Dot(A,A));
    }
    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;
    }
    
    bool operator == (const Point& a,const Point& b)
    {
        return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
    }
    bool OnSegment(Point p,Point a1,Point a2)
    {
        return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0;
    }
    int cmp(const Point &a,const Point &b)
    {
        if(a.x<b.x) return 1;
        else if(a.x==b.x)
        {
            if(a.y<b.y) return 1;
            else return 0;
        }
        else return 0;
    }
    int ConvexHull(Point *p,int n,Point *ch)
    {
        sort(p,p+n,cmp);
        n=unique(p,p+n)-p;
        int m=0;
        for(int i=0;i<n;i++)
        {
            while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]) <= 0)
                m--;
            ch[m++]=p[i];
        }
        int k=m;
        for(int i=n-2;i>=0;i--)
        {
            while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]) <= 0)
                m--;
            ch[m++]=p[i];
        }
        if(n>1) m--;
        return m;
    }
    int isPointInPolygon(Point p,Point* poly,int n)
    {
        int wn=0;
        for(int i=0;i<n;i++)
        {
            Point& p1=poly[i],p2=poly[(i+1)%n];
            if(p==p1 || p==p2 || OnSegment(p,p1,p2)) return -1;
            int k=dcmp(Cross(p2-p1,p-p1));
            int d1=dcmp(p1.y-p.y);
            int d2=dcmp(p2.y-p.y);
            if(k>0 && d1<=0 && d2>0) ++wn;
            if(k<0 && d2<=0 && d1>0) --wn;
        }
        if(wn) return 1;
        return 0;
    }
    bool SegmentProperIntersect(Point a1,Point a2,Point b1,Point b2)
    
    {
        double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
        double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
        return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
    }
    
    //Point read()
    //{
    //    Point p;
    //    scanf("%d%d",&p.x,&p.y);
    //    return p;
    //}
    void read(Point& p)
    {
        scanf("%d%d",&p.x,&p.y);
    }
    bool solve()
    {
        Point chr[maxn],chb[maxn];
        int mr,mb;
        mr=ConvexHull(red,n,chr);
        mb=ConvexHull(blue,m,chb);
        for(int i=0;i<mr;i++)
            if(isPointInPolygon(chr[i],chb,mb))
               return 0;
        for(int i=0;i<mb;i++)
            if(isPointInPolygon(chb[i],chr,mr))
               return 0;
        for(int i=0;i<mr;i++)
            for(int j=0;j<mb;j++)
            {
                if(SegmentProperIntersect(chr[i],chr[(i+1)%mr],chb[j],chb[(j+1)%mb]))
                   return 0;
            }
        return 1;
    }
    int main()
    {
      //  freopen("ine.cpp","r",stdin);
        while(scanf("%d%d",&n,&m) && n && m)
        {
            for(int i=0;i<n;i++)
                read(red[i]);//red[i]=read();//
            for(int i=0;i<m;i++)
                read(blue[i]);//blue[i]=read();//
            if(solve()) printf("Yes
    ");
            else printf("No
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/54zyq/p/3234305.html
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