<h1 id="heading">题目</h1>
<p>{% fold 点击显/隐题目 %}</p>
<div class="oj"><div class="part" title="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. 
<p>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.</p>
</div><div class="part" title="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).
</div><div class="part" title="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".
</div><div class="samp"><div class="clear"></div><div class="input part" title="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
</div><div class="output part" title="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
</div><div class="clear"></div></div></div>
{% endfold %}

<h1 id="heading-1">题解</h1>
<p>判断两条<strong>直线</strong>的关系<br>
关系有:</p>
<ul>
<li>重合</li>
<li>相交</li>
<li>不相交<br>
其中,对于相交还需要求出交点</li>
</ul>
<p>可以使用矩阵解交点方程<br>
根据不同情况返回不同值</p>
<h1 id="heading-2">代码</h1>
<p>{% fold 点击显/隐代码 %}```cpp Intersecting Lines <a href="https://github.com/OhYee/sourcecode/tree/master/ACM">https://github.com/OhYee/sourcecode/tree/master/ACM</a> 代码备份<br>
#include <algorithm><br>
#include <cmath><br>
#include <complex><br>
#include <cstdio><br>
#include <vector><br>
using namespace std;</p>
<p>#define Log(format, ...) // printf(format, ##<strong>VA_ARGS</strong>)</p>
<p>/* 计算几何模板 */<br>
const double eps = 1e-8;</p>
<p>typedef complex<double> Vector;<br>
typedef Vector Point;<br>
struct Segment {<br>
Point a, b;<br>
Segment() {}<br>
Segment(Point _a, Point _b) : a(_a), b(_b) {}<br>
Vector toVector() { return b - a; }<br>
};</p>
<p>inline Point read_Point() {<br>
double x, y;<br>
scanf(&quot;%lf%lf&quot;, &amp;x, &amp;y);<br>
return Point(x, y);<br>
}<br>
inline int sgn(const double &amp;x) {<br>
if (fabs(x) &lt; eps)<br>
return 0;<br>
return x &gt; 0 ? 1 : -1;<br>
}<br>
inline bool operator==(const Vector &amp;vec1, const Vector &amp;vec2) {<br>
return sgn(vec1.real() - vec2.real()) == 0 &amp;&amp;<br>
sgn(vec1.imag() - vec2.imag()) == 0;<br>
}<br>
inline bool operator!=(const Vector &amp;vec1, const Vector &amp;vec2) {<br>
return !(vec1 == vec2);<br>
}</p>
<p>inline int Cross(const Vector &amp;a, const Vector &amp;b) {<br>
return a.real() * b.imag() - a.imag() * b.real();<br>
}<br>
inline int Dot(const Vector &amp;a, const Vector &amp;b) {<br>
return a.real() * b.real() + a.imag() * b.imag();<br>
}<br>
inline double Distance(const Point &amp;a, const Point &amp;b) { return abs(a - b); }<br>
inline int Point_Segment(const Vector &amp;p, const Segment &amp;L) {<br>
return sgn(Cross(L.b - L.a, p - L.a));<br>
}<br>
inline bool Segment_Segment(const Segment &amp;L1, const Segment &amp;L2) {<br>
if (sgn(Distance(L1.a, L1.b)) == 0 || sgn(Distance(L2.a, L2.b)) == 0)<br>
return 0;<br>
return (Point_Segment(L1.a, L2) * Point_Segment(L1.b, L2) &lt;= 0) &amp;&amp;<br>
(Point_Segment(L2.a, L1) * Point_Segment(L2.b, L1) &lt;= 0);<br>
}<br>
inline int Segment_Line(const Segment &amp;Seg, const Segment &amp;Line) {<br>
Log(&quot;(%.f,%.f)-&gt;(%.f,%.f)   (%.f,%.f)-&gt;(%.f,%.f)\n&quot;, Seg.a.real(),<br>
Seg.a.imag(), Seg.b.real(), Seg.b.imag(), Line.a.real(), Line.a.imag(),<br>
Line.b.real(), Line.b.imag());<br>
if (sgn(Distance(Line.a, Line.b)) == 0)<br>
return 0;<br>
return Point_Segment(Seg.a, Line) * Point_Segment(Seg.b, Line) &lt;= 0;<br>
}<br>
inline int getPoint(const Segment L1, const Segment L2, Point &amp;p) {<br>
double a = L1.b.real() - L1.a.real();<br>
double b = L2.b.real() - L2.a.real();<br>
double c = L1.b.imag() - L1.a.imag();<br>
double d = L2.b.imag() - L2.a.imag();<br>
double f = a * d - b * c;</p>
<pre><code>// 平行或重叠
if (sgn(f) == 0)
    return -1;

double g = L2.b.real() - L1.a.real();
double h = L2.b.imag() - L1.a.imag();
double t = (d * g - b * h) / f;
double s = (-c * g + a * h) / f;

p = Point(L1.a.real() + t * a, L1.a.imag() + t * c);
// 在延长线上
if (t &lt; 0 || t &gt; 1 || s &lt; 0 || s &gt; 1)
    return 0;

// 线段相交
return 1;
</code></pre>
<p>}</p>
<p>int main() {<br>
int T;<br>
scanf(&quot;%d&quot;, &amp;T);<br>
printf(&quot;INTERSECTING LINES OUTPUT\n&quot;);<br>
while (T--) {<br>
Point L1a = read_Point();<br>
Point L1b = read_Point();<br>
Point L2a = read_Point();<br>
Point L2b = read_Point();</p>
<pre><code>    Segment L1 = Segment(L1a, L1b);
    Segment L2 = Segment(L2a, L2b);
    Point p;
    int state = getPoint(L1, L2, p);

    if (state &gt;= 0) {
        printf(&quot;POINT %.2f %.2f\n&quot;, p.real(), p.imag());
    } else if (Point_Segment(L1.a, L2) == 0) {
        printf(&quot;LINE\n&quot;);
    } else {
        printf(&quot;NONE\n&quot;);
    }
}
printf(&quot;END OF OUTPUT\n&quot;);
return 0;
</code></pre>
<p>}</p>
<pre style="background-color:#fff">{% endfold %}</pre>

POJ 1269.Intersecting Lines