poj1269计算几何直线和直线的关系

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
很简单直接暴力分类，类别也不是很多，有一个坑点就是double型的0乘负数会变成负0，太坑了！！
这里放一下测试代码

#include<map>
#include<set>
#include<list>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
 
using namespace std;
 
const int N=,maxn=,inf=0x3f3f3f3f3f;
 
int main()
{
    double x=0.0,y=x*(-);
    printf("%.2f\n",y);
    if(y==)y=fabs(y);
    printf("%.2f\n",y);
    return ;
}

#include<map>
#include<set>
#include<list>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
 
using namespace std;
 
const double eps=1e-;
const int N=,maxn=,inf=0x3f3f3f3f;
 
struct point{
    int x,y;
};
struct line{
   point a,b;
}l[N];
 
int main()
{
    int t;
    double x1,y1,x2,y2,x3,y3,x4,y4;
    cin>>t;
    cout<<"INTERSECTING LINES OUTPUT"<<endl;
    while(t--){
        cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
        if((y4-y3)*(x2-x1)==(y2-y1)*(x4-x3))
        {
            if((y3-y1)*(x2-x1)!=(y2-y1)*(x3-x1))
                cout<<"NONE"<<endl;
            else
                cout<<"LINE"<<endl;
        }
        else
        {
            double x,y;
            if(x2==x1)
            {
                x=x1;
                y=y3+(x-x3)*(y4-y3)/(x4-x3);
            }
            else if(x3==x4)
            {
                x=x3;
                y=y1+(x-x1)*(y2-y1)/(x2-x1);
            }
            else
            {
                x=(y3-y1+x1*(y2-y1)/(x2-x1)-x3*(y4-y3)/(x4-x3))/((y2-y1)/(x2-x1)-(y4-y3)/(x4-x3));
                y=(x-x1)*(y2-y1)/(x2-x1)+y1;
            }
            if(x==)x=fabs(x);
            if(y==)y=fabs(y);
            printf("POINT %.2f %.2f\n",x,y);
        }
    }
    cout<<"END OF OUTPUT"<<endl;
    return ;
}

又看了一下网上的题解发现有更简单的叉积判断

首先判断斜率是非相同还是用公式直接来(x4-x3)*(y2-y1)==(y4-y3)*(x2-x1)

然后用叉积(x2-x1)*(y3-y1)==(y2-y1)*(x3-x1)判断x3是不是在x1，x2这条线上是的话就是LINE，否则就是NONE

最后叉积计算交点：

设交点（x0，y0）

(x2-x1)*(y0-y1)-(y2-y1)*(x0-x1)=0;

(x4-x3)*(y0-y3)-(y4-y3)*(x0-x3)=0;

化简可得：

(y1-y2)*x0+(x2-x1)*y0+x1*y2-x2*y1=0;

(y3-y4)*x0+(x4-x3)*y0+x3*y4-x4*y3=0;

建立二元一次方程：

a1*x0+b1*y0+c1=0;

a2*x0+b2*y0+c2=0;

解得：

x0=(c2*b1-c1*b2)/(b2*a1-b1*a2);

y0=(a2*c1-a1*c2)/(b2*a1-b1*a2);

带入就好了，以下是新方法 的ac代码：

#include<map>
#include<set>
#include<list>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
 
using namespace std;
 
const double eps=1e-;
const int N=,maxn=,inf=0x3f3f3f3f;
 
struct point{
    double x,y;
};
struct line{
   point a,b;
}l[N];
 
int main()
{
    int t;
    double x1,x2,x3,x4,y1,y2,y3,y4;
    cin>>t;
    cout<<"INTERSECTING LINES OUTPUT"<<endl;
    while(t--){
        cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
        if((x4-x3)*(y2-y1)==(y4-y3)*(x2-x1))//斜率判断
        {
            if((x2-x1)*(y3-y1)==(y2-y1)*(x3-x1))cout<<"LINE"<<endl;//用叉积判断共线
            else cout<<"NONE"<<endl;
        }
        else
        {
            double a1=y1-y2,b1=x2-x1,c1=x1*y2-x2*y1;
            double a2=y3-y4,b2=x4-x3,c2=x3*y4-x4*y3;
            double x=(c2*b1-c1*b2)/(b2*a1-b1*a2);
            double y=(a2*c1-a1*c2)/(b2*a1-b1*a2);
            printf("POINT %.2f %.2f\n",x,y);
        }
    }
    cout<<"END OF OUTPUT"<<endl;
    return ;
}