C How to calculate whether two regions intersect

Both regions should have x and y coordinates, right? It is enough to determine whether the upper left point and the lower left point of the second area are in the first area, if they are, it means that there is anxiety, if not, there is no intersection.

For example, xb > xa, and rectangle a is on the left side of rectangle b, then use the rightmost line of a to compare the leftmost line of b, and if there is an intersection point, it means that the rectangle has intersected (so that only one line intersects!). You can also take the center point as the center of the circle, and take the farthest point as the radius, and compare whether the two are too far apart, and if they are too far away, they don't even need to intersect, and it is more practical to call them repeatedly).

The above edge should be compared to the long side (the edge of the point farther away).

Do you have to write** Don't want to write.

Four points make a b c d

First, find the one with a small distance from a and b to the straight line cd, let it be l, or judge the angle and use the slope to find one

Just judge which is bigger, L and D(1+Sin), L is big and doesn't intersect, it's not a waste of time to have an open root (find the distance), and it's not difficult to write a tan sin.

It is conceivable to hold the center of the rectangle and rotate it, and in the most extreme case, the two rectangles are exactly half the sum of their diagonal distances.

Separate each edge of rectangles A and B. Determine whether the two rectangles intersect by looking at whether the 4 sides of A and the 4 sides of B intersect.

See if it helps you.

This uses vectors as containers.

Then compare it yourself with a loop.

It is a good topic for exercising basic procedural skills.

Find out the same content to generate a linked list and put it in head1.

Let's assume that head1 and head2 are already sorted by data.

If the data of head1 is smaller than that of head2, the linked list of head1 deletes the node and points to the next node for comparison.

If the data of head1 is greater than that of head2, the linked list of head2 points to the next node and the comparison continues.

If the data of the two nodes is the same, keep the node of head1 (find out the same content), head1, and head2 all point to the next node and continue to compare.

The data structure C language version, look at the numerical position.

A similar linked list algorithm. Sorted, isn't it?

Hello, the specific data type is not given in the question, so the following is given pseudo** as an idea.

If you know the specific data type, there may be an off-the-shelf solution in the C++ library) class a as a collection element.

class b as a collection.

Method b2} that returns a union with another set