There are two side rectangles that are parallel to the axes and they intersect. We know:
(x1, y1) - the left bottom point of the first rectangle
(x2, y2) - the upper right point of the first rectangle
(x3, y3) - the left bottom point of the second rectangle
(x4, y4) - the upper right point of the second rectangle
And you need to find the area of their intersection. But they can cross from different sides.
Although the question is simple, I'll leave it as a snippet crib:
#include <algorithm> /* x1, y1 - левая нижняя точка первого прямоугольника x2, y2 - правая верхняя точка первого прямоугольника x3, y3 - левая нижняя точка второго прямоугольника x4, y4 - правая верхняя точка второго прямоугольника */ int f(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) { int left = std::max(x1, x3); int top = std::min(y2, y4); int right = std::min(x2, x4); int bottom = std::max(y1, y3); int width = right - left; int height = top - bottom; if ((width < 0) || (height < 0)) return 0; return width * height; } Based on the text of the question, I believe that the coordinates grow from the lower left corner (if Y grows from top to bottom, amend the corresponding ones).
The idea is simple, illustrated in the picture (for the width, for the height is the same):
max . - IgorSource: https://ru.stackoverflow.com/questions/758529/
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с.left = max(a.left, b.left);,с.top = max(a.top, b.top);,c.right = min(a.right, b.right );,c.bottom = min(a.bottom, b.bottom );(well, another check that the non-negative dimensions turned out, in case they do not overlap). - Vladimir Gamalyan