blocks|key|2121924|text|可以通过求解图的所有边对的交点方程来计算交点:+/frac{x+-+a}{b+-+a}+=+/frac{x+-+c}{d+-+c}|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2121925|你以这种方式获得的观点可以放在帕雷培德的两侧，尽管它们不能。你必须检查你通过求解方程得到的交点是否在图形的两侧。如果是这样的话，你可以计算出延伸到这两个图形内部的图形的边长，并通过求出它们的倍数来计算相交的平方。|2121926|我想我的方法听起来有点复杂，但这是我第一个想到的。|2121927|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

You can calculate the intersection points by solving intersection equations for all pairs of sides of the figures:
/frac{x - a}{b - a} = /frac{x - c}{d - c}

The points that you obtain in such a fashion can lie on the sides of the paralelepide, though they must not. You have to check whether the intersection points you obtained by solving the equations lie on the sides of the figure or not. If they do, you can calculate the length of the sides of the figure that stretch out into the inside of the both figures, and calculate the square of the intersection by taking their multiple.

I guess my method sounds a bit over-complicated, but this is the first thought that came to my mind.

blocks|key|2121959|text|因为你说矩形可能不对齐，所以可能的答案可能是无意义的，一个点，一个线段，或一个3-8边的多边形。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2121960|做这个2d+boolean操作的通常方法是选择边的逆时针顺序，然后评估临界点(交叉点或角)之间的边缘段。在每个交点，您切换到第一个矩形的边缘段到第二个矩形的边缘，或者相反。您总是选择前一段左边的段。|offset|length|style|CODE|2121961|​|2121962|📷|atomic|2121963|2121964|有很多细节，但基本的算法是找出所有的交叉口，并在它们的边缘与适当的数据结构排序。选择一个交叉口(如果有)，并选择一个从该交叉口引向的线段。找到所选开始段左侧的另一个矩形段。在图中，我们选择相交a(箭头所指示的方向)上的绿色段作为参考段。右边的另一个矩形段是从a到b的段。使用它作为下一个参考段，并在其左边选择一个绿色段。这是从b到c的段，用同样的方式查找段cd。下一段从d到拐角处，所以拐角处也在交集的顶点列表中。从玉米里我们回到了一个。|2121965|若要每次选择左侧，您可以使用方向矢量坐标的确定值来确定满足的边缘。如果有序对有向边的行列式是正数，你就走对了路。|2121966|现在，您有了交集多边形的顶点，您可以使用http://en.wikipedia.org/wiki/Polygon#Area_and_centroid来获取该区域。|2121967|我要留给你们的一些细节如下：|2121968|如果一个角与另一个三角形的边或顶点重合怎么办？|unordered-list-item|2121969|如果没有交叉口怎么办？(一个矩形在另一个内，或者它们是不相交的-你可以用点中多边形检查来找出这一点。见维基百科关于多边形的文章。|2121970|如果交点是一个点或一个段呢？|2121971|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/b0e189379cee943cbd50e9efdf7cb9cd.png|imageAlt|1|LINK|MUTABLE|url|http://en.wikipedia.org/wiki/Polygon#Area_and_centroid|2|http://en.wikipedia.org/wiki/Polygon^0|0|3|A|0|0|0|1|0|0|0|0|0|K|1I|1|0|0|0|1F|C|2|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1J|8|@$D|1K|E|1L|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1M|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|1N|8|@]|9|@$D|1O|E|1P|1|1Q]]|A|$]]|$1|M|3|I|5|6|7|1R|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1S|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1T|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1U|8|@]|9|@$D|1V|E|1W|1|1X]]|A|$]]|$1|T|3|U|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|W|5|X|7|1Z|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|X|7|20|8|@]|9|@$D|21|E|22|1|23]]|A|$]]|$1|10|3|11|5|X|7|24|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|25|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19|1A|-4]]|1B|$5|1C|16|1D|A|$1E|1F]]|1G|$5|1C|16|1D|A|$1E|1H]]]]

Since you stated that the rectangles may not be aligned, possible answers may be nothing, a point, a line segment, or a polygon with 3-8 sides.

The usual way to do this <code>2d boolean</code> operation is to choose a counterclockwise ordering of the edges, and then evaluate edge segments between critical points (intersections or corners). At each intersection you switch between an edge segment of the first rectangle to an edge of the second, or visa-versa. You always pick the segment to the left of the previous segment.

<img src="https://i.stack.imgur.com/Q2zfd.png" alt="enter image description here">

There are LOTS of details, but the basic algorithm is to find all intersections and order them on their edges with an appropriate data structure. Choose an intersection (if there is one) and choose a line segment leading away from that intersection. Find the segment of the other rectangle to the left of the chosen starting segment. In the picture, we choose the green segment on intersection a (in the direction indicated by the arrow) as the reference segment. The segment of the other rectangle that is to the right, is the segment from a to b. Use that as the next reference segment, and choose a green segment to the left of it. That's the segment from b to c. Find segment cd the same way. The next segment is from d to the corner, so the corner is in the vertex list for the intersection as well. From the corn we get back to a.

To choose the left side each time, you use the determinate of the coordinates of the direction vectors for the edges that meet. If the determinant for the ordered pair of directed edges is positive, you're going the right way.

Now that you have the vertices of the intersection polygon, you can use the <a href="http://en.wikipedia.org/wiki/Polygon#Area_and_centroid" rel="nofollow noreferrer">surveyor's formula</a> to get the area.

Some of the details that I'm leaving to you are:

<ul>
<li>What if a corner is coincident to to an edge or vertex of the other triangle?</li>
<li>What if there are no intersections? (one rectangle is inside the other, or they are disjoint--you can use point-in-polygon checks to figure this out. See the <a href="http://en.wikipedia.org/wiki/Polygon" rel="nofollow noreferrer">Wikipedia article on polygons</a>.</li>
<li>What if the intersection is a single point or a segment?</li>
</ul>

blocks|key|2122023|text|您可能会发现另一种有趣的方法，但在这种情况下可能不适用，那就是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2122024|确定包含两个给定矩形的最小矩形(其边与坐标轴平行)，让我们将新矩形称为包围框。|ordered-list-item|2122025|选择一个在边界框中的随机点，并检查它是否位于两个矩形中。|2122026|重复步骤2，只要你愿意(这取决于你想要的结果的精度)，并且有两个计数器，一个用来跟踪两个矩形内的点数，另一个是步骤2的重复次数。|2122027|最后的解是包围框的面积乘以两个矩形内的点数，然后除以步骤2的重复次数，或以一种公式的形式：
intersection_area+=+bounding_box_area+*+num_of_dots_inside_both+/+num_of_repetitions|2122028|当然，当重复次数更多时，结果会更精确。顺便说一句，这种方法叫做蒙特卡罗法。|2122029|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|Q|8|@]|9|@]|A|$]]|$1|G|3|H|5|D|7|R|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|S|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|T|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|N|$]]

There is another way that you may find interesting, but maybe not applicable in this case, and that is:

<ol>
<li>determine the minimum rectangle( whose sides are parallel to
coordinate axes ) that contains both of the given rectangles, lets
call that new one a bounding box.</li>
<li>pick a random dot that is in the bounding box and check whether it is in both rectangles or not</li>
<li>repeat step 2 for as long as you want( it depends on the precision you want for your result ), and have two counters, one to
keep track of the number of dots inside both of the rectangles, and
the other which is the number of repetitions of step 2</li>
<li>the final solution is the area of the bounding box multiplied by the number of dots inside both rectangles and then divided by number
of repetitions of step 2, or in a form of a formula:

intersection_area = bounding_box_area * num_of_dots_inside_both /
num_of_repetitions</li>
</ol>

The result will, of course, be more precise when the number of repetitions is larger. By the way, this method is called Monte Carlo method.

blocks|key|2122088|text|用三角形而不是矩形来思考这个问题可能会有所帮助。在空间中找出一个三角形的面积，给出三个点，相对来说是直线的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2122089|您可以通过将矩形区域减去三角形的面积之和来找到相交区域，如下图所示。|2122090|​|2122091|📷|atomic|offset|length|2122092|2122093|从本质上说，它变成了一个三角测量问题。|2122094|有一个很好的介绍性的这里和一些关于数据结构和算法的指针。|2122095|编辑：|2122096|有一些自由三角网库可以重用。|2122097|如果你知道两个三角形的面积，你可以找到矩形合并的总面积，所以相交就是两个矩形的总面积--联合区域。|2122098|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/7acab020882bdc015c662af7a77dcfcd.png|imageAlt|1|LINK|MUTABLE|url|http://mathworld.wolfram.com/Triangulation.html|2|http://www.cs.ucsb.edu/~suri/cs235/Triangulation.pdf|3|http://vterrain.org/Implementation/Libs/triangulate.html^0|0|0|0|0|1|0|0|0|C|6|1|0|A|2|2|0|0|3|6|3|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1E|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1F|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|1G|8|@]|9|@$I|1H|J|1I|1|1J]]|A|$]]|$1|K|3|E|5|6|7|1K|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1L|8|@]|9|@$I|1M|J|1N|1|1O]]|A|$]]|$1|N|3|O|5|6|7|1P|8|@]|9|@$I|1Q|J|1R|1|1S]]|A|$]]|$1|P|3|Q|5|6|7|1T|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1U|8|@]|9|@$I|1V|J|1W|1|1X]]|A|$]]|$1|T|3|U|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12|13|-4]]|14|$5|15|Z|16|A|$17|18]]|19|$5|15|Z|16|A|$17|1A]]|1B|$5|15|Z|16|A|$17|1C]]]]

It may help to think of the problem in terms of triangles instead of rectangles. Finding the area of a triangle given three points in space is relatively straight forward.

You can find the intersecting area by subtracting the rectangle area by the sum of the areas of the triangles as shown in the image below.

<img src="https://i.stack.imgur.com/OyGmH.png" alt="Triangulation">

Essentially it becomes a <a href="http://mathworld.wolfram.com/Triangulation.html" rel="nofollow noreferrer">triangulation problem</a>.

There is a good intro <a href="http://www.cs.ucsb.edu/~suri/cs235/Triangulation.pdf" rel="nofollow noreferrer">here</a> with some pointers on data structures and algorithms.

EDIT:

There are some <a href="http://vterrain.org/Implementation/Libs/triangulate.html" rel="nofollow noreferrer">free triangulation libraries</a> that you could reuse. 

If you know the area of the two triangles you are starting off with you can find the total area of the union of the rectangles, so the intersection would be the total area of both rectangles - the union area.

blocks|key|136505|text|如果您碰巧使用了Qt，那么可以将该交集计算为QPolygonF交集。大致如此：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|136506|QPolygonF+p1,p2,+intersected;
p1+<<+QPointF(r1x1,r1y1)+<<+...+<<+QPointF(r1x4,+r1y4);
p2+<<+QPointF(r2x1,r2y2)+<<+...+<<+QPointF(r2x4,+r2y4);
intersected+=+p1.intersected(p2);

float+area+=+polyArea(intersected);+//+see+code+block+below|code-block|syntax|javascript|136507|(r1+=矩形1，r2+=矩形2，4个分别为x和y坐标)。|136508|现在计算面积(使用前面提到的鞋带公式)：|136509|inline+float+polyArea(const+QPolygonF&+p)
{
++++//https://en.wikipedia.org/wiki/Polygon#Area_and_centroid
++++const+int+n+=+p.size();
++++float+area+=+0.0;
++++for+(int+i=0;+i<n;+i%2B%2B)
++++{
++++++++area+%2B=+p[i].x()*p[(i%2B1)%25n].y()+-+p[(i%2B1)%25n].x()*p[i].y();
++++}
++++if+(area+<+0)
++++++++return+-0.5*area;
++++else
++++++++return+0.5*area;
}|136510|我这里的代码:公共域|136511|entityMap|0|LINK|mutability|MUTABLE|url|http://https//en.wikipedia.org/wiki/Polygon#Area_and_centroid^0|8|2|M|9|0|0|0|E|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|16|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|17|8|@]|E|@]|F|$]]|$1|N|3|O|5|6|7|18|8|@]|E|@$9|19|A|1A|1|1B]]|F|$]]|$1|P|3|Q|5|I|7|1C|8|@]|E|@]|F|$J|K]]|$1|R|3|S|5|6|7|1D|8|@]|E|@]|F|$]]|$1|T|3|-4|5|6|7|1E|8|@]|E|@]|F|$]]]|U|$V|$5|W|X|Y|F|$Z|10]]]]

If you happen to use Qt then the intersection can be computed as <code>QPolygonF</code> intersection.
Roughly as so:

<pre><code>QPolygonF p1,p2, intersected;
p1 &lt;&lt; QPointF(r1x1,r1y1) &lt;&lt; ... &lt;&lt; QPointF(r1x4, r1y4);
p2 &lt;&lt; QPointF(r2x1,r2y2) &lt;&lt; ... &lt;&lt; QPointF(r2x4, r2y4);
intersected = p1.intersected(p2);

float area = polyArea(intersected); // see code block below
</code></pre>

(r1 = rectangle 1, r2 = rectangle 2, with 4 respective x and y coordinates). 

Now compute the area (using the already mentioned <a href="http:////https://en.wikipedia.org/wiki/Polygon#Area_and_centroid" rel="nofollow">Shoelace formula</a>):

<pre><code>inline float polyArea(const QPolygonF&amp; p)
{
 //https://en.wikipedia.org/wiki/Polygon#Area_and_centroid
 const int n = p.size();
 float area = 0.0;
 for (int i=0; i&lt;n; i++)
 {
 area += p[i].x()*p[(i+1)%n].y() - p[(i+1)%n].x()*p[i].y();
 }
 if (area &lt; 0)
 return -0.5*area;
 else
 return 0.5*area;
}
</code></pre>

My code here: public domain

blocks|key|136515|text|也许你需要使用opencv。使用fillPoly()函数生成一个矩形。确保填充矩形为白色(255,255,255)。然后使用copyTo()函数，您将得到重叠区域。然后检查每个像素的值，如果它是白色的，那么%2B1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|136516|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Maybe you need to use opencv. Use the fillPoly() function to generate a rectangle. Make sure the fill rectangle is white (255, 255, 255). Then use the copyTo() function and you will get the overlap area. Then check the value of each pixel , if it is white then +1.

Below are 2 rectangles. Given the coordinates of the rectangle vertices - (x1, y1)...(x8, y8), how can the area of the overlapping region (white in the figure below) be caclulated?
 Note that:

<ol>
<li>Coordinates of points might be any</li>
<li>Rectangles may or may not overlap</li>
<li>Assume area is 0 when rectangles don't overlap, or they overlap at point or line.</li>
<li>If one rectangle is inside the other, then calculate area of smaller rectangle.</li>
</ol>

<img src="https://i.stack.imgur.com/stTc5.jpg" alt="enter image description here">

Area of rectangle-rectangle intersection

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面是两个矩形。给定矩形顶点的坐标- (x1，y1).(x8，y8)，如何计算重叠区域(下图中为白色)的面积？请注意：点的坐标可能是矩形可能重叠，也可能不重叠假设矩形不重叠或在点或线重叠时面积为0。如果一个矩形位于另一个内，则计算较小矩形的面积。​​

问矩形-矩形相交面积
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