poj 1269

最新推荐文章于 2020-10-26 19:25:48 发布

weixin_30790841

最新推荐文章于 2020-10-26 19:25:48 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zwx7616/p/11189746.html

博客主要提及相交求交点，给出第二个易记忆公式，并以 poj 1269 intersecting lines 为例，还给出了转载链接。

相交求交点：

第二个公式容易记忆：（以第二个公式为模板）

Point res=b.s;
t=((e-s)^(s-b.s))/((e-s)^(b.e-b.s));
res.x+=(e.x-s.x)*t;
res.y+=(e.y-s.y)*t;
return res;

例题：

poj 1269

intersecting lines

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int cmp(double x)
{
    if(fabs(x)<=1e-8)return 0;
    if(x<0)return -1;
    return 1;
}
struct Point
{
    double x,y;
    Point (){};
    Point (double _x,double _y)
    {
        x=_x,y=_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);
    }
};

struct Line
{
    Point s,e;
    Line(){};
    Line(Point _s,Point _e)
    {
        s=_s,e=_e;
    }
    Point operator &(const Line &b)const{
    Point res=b.s;
    if(cmp((e-s)^(b.e-b.s))==0)
    {
        if(cmp((e-s)^(e-b.e))==0)
        {
            cout<<"LINE"<<endl;
            return Point(0,0);
        }
        else
            {
                cout<<"NONE"<<endl;
                return Point(0,0);
            }
    }
    double t=((e-s)^(s-b.s))/((e-s)^(b.e-b.s));
    res.x+=(b.e.x-b.s.x)*t;
    res.y+=(b.e.y-b.s.y)*t;
    return res;
    }
};


int main ()
{
    int n;
    cin>>n;
    Line line1,line2;
    Point point ;
    double x1,x2,x3,x4,y1,y2,y3,y4;
     cout<<"INTERSECTING LINES OUTPUT"<<endl;
    while(n--)
    {
        cin>>x1>>y1>>x2>>y2;
        line1=Line(Point(x1,y1),Point(x2,y2));
        cin>>x3>>y3>>x4>>y4;
        line2=Line(Point(x3,y3),Point(x4,y4));
        point=line1&line2;
        if(point.x!=0||point.y!=0)
          printf("POINT %.2f %.2f\n",point.x,point.y);
    }
    cout<<"END OF OUTPUT"<<endl;
    return 0;
}