POJ 1269

Time Limit: 1000MS Memory Limit: 10000K

Description

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.

Input

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

Output

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

题意：

给出四个点的坐标，分别用来表示两条直线，求两条直线的关系是共线还是平行还是相交，若是相交则给出交点坐标（保留小数点后两位）。

题解：

直接套用模板中的点到直线距离以及求直线交点的函数模板即可。

AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
//--------------------------------------计算几何模板 - st--------------------------------------

const double eps = 1e-6;
const double M_PI = 3.14159265358979323846;

struct Point{
    double x,y;
    Point(double tx=0,double ty=0):x(tx),y(ty){}
};
typedef Point Vctor;

//向量的加减乘除
Vctor operator + (Vctor A,Vctor B){return Vctor(A.x+B.x,A.y+B.y);}
Vctor operator - (Point A,Point B){return Vctor(A.x-B.x,A.y-B.y);}
Vctor operator * (Vctor A,double p){return Vctor(A.x*p,A.y*p);}
Vctor operator / (Vctor A,double p){return Vctor(A.x/p,A.y/p);}
bool operator < (Point A,Point B){return A.x < B.x || (A.x == B.x && A.y < B.y);}

struct Line{
    Point p;
    Vctor v;
    Line(Point p=Point(0,0),Vctor v=Vctor(0,0)):p(p),v(v){}
    Point point(double t){return p + v*t;} //获得直线上的距离p点t个单位长度的点
};
struct Circle{
    Point c;
    double r;
    Circle(Point tc=Point(0,0),double tr=0):c(tc),r(tr){}
    Point point(double a){return Point( c.x + cos(a)*r , c.y + sin(a)*r);}
};

int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return (x<0)?(-1):(1);
}
bool operator == (Point A,Point B){return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0;}

//向量的点积，长度，夹角
double Dot(Vctor A,Vctor B){return A.x*B.x+A.y*B.y;}
double Length(Vctor A){return sqrt(Dot(A,A));}
double Angle(Vctor A,Vctor B){return acos(Dot(A,B)/Length(A)/Length(B));}

//叉积，三角形面积
double Cross(Vctor A,Vctor B){return A.x*B.y-A.y*B.x;}
double TriangleArea(Point A,Point B,Point C){return Cross(B-A,C-A);}

//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）
Vctor Rotate(Vctor A,double rad){return Vctor( A.x*cos(rad) - A.y*sin(rad) , A.x*sin(rad) + A.y*cos(rad) );}
Vctor Normal(Vctor A)
{
    double L = Length(A);
    return Vctor(-A.y/L, A.x/L);
}

//直线的交点
Point getLineIntersection(Line L1,Line L2)
{
    Vctor u = L1.p-L2.p;
    double t = Cross(L2.v,u)/Cross(L1.v,L2.v);
    return L1.p + L1.v*t;
}

//点到直线的距离
double DistanceToLine(Point P,Line L)
{
    return fabs(Cross(P-L.p,L.v))/Length(L.v);
}

//--------------------------------------计算几何模板 - ed--------------------------------------
double x[5],y[5];
Line L1,L2;
int CheckTwoLines(Line L1,Line L2)//1-共线，2-平行，3-相交
{
    if(dcmp( L1.v.x * L2.v.y - L2.v.x * L1.v.y ) == 0 ) //两直线的方向向量平行
    {
        if( DistanceToLine(L1.p,L2) < eps ) //直线1上的点也在直线2上
            return 1;
        else
            return 2;
    }
    else return 3;
}
int main()
{
    int t;
    scanf("%d",&t);
    printf("INTERSECTING LINES OUTPUT
");
    while(t--)
    {
        for(int i=1;i<=4;i++) scanf("%lf%lf",x+i,y+i);
        L1=Line(Point(x[1],y[1]),Point(x[2],y[2])-Point(x[1],y[1]));
        L2=Line(Point(x[3],y[3]),Point(x[4],y[4])-Point(x[3],y[3]));

        int relation=CheckTwoLines(L1,L2);
        if(relation==1)
            printf("LINE
");
        else if(relation==2)
            printf("NONE
");
        else
        {
            Point ans=getLineIntersection(L1,L2);
            printf("POINT %.2f %.2f
",ans.x,ans.y);
                //此处使用G++时用 %.2f 输出，使用C++时用 %.2lf 输出
        }
    }
    printf("END OF OUTPUT
");
}