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    <p>An efficient way of computing this area is to use a sweep algorithm. Let us assume that we sweep a vertical line L(x) through the union of rectangles U: </p> <ul> <li>first of all, you need to build an event queue Q, which is, in this case, the ordered list of all x-coordinates (left and right) of the rectangles. </li> <li>during the sweep, you should maintain a 1D datastructure, which should give you the total length of the intersection of L(x) and U. The important thing is that this length is constant between two consecutive events q and q' of Q. So, if l(q) denotes the total length of L(q+) (i.e. L just on the rightside of q) intersected with U, the area swept by L between events q and q' is exactly l(q)*(q' - q).</li> <li>you just have to sum up all these swept areas to get the total one.</li> </ul> <p>We still have to solve the 1D problem. You want a 1D structure, which computes dynamically a union of (vertical) segments. By dynamically, I mean that you sometimes add a new segment, and sometimes remove one. </p> <p>I already detailed in my answer to this <a href="https://stackoverflow.com/questions/149577/need-an-algorithm-for-collapsing-netblock-ranges-into-lists-of-superset-ranges">collapsing ranges question</a> how to do it in a static way (which is in fact a 1D sweep). So if you want something simple, you can directly apply that (by recomputing the union for each event). If you want something more efficient, you just need to adapt it a bit:</p> <ul> <li>assuming that you know the union of segments S<sub>1</sub>...S<sub>n</sub> consists of disjoints segments D<sub>1</sub>...D<sub>k</sub>. Adding S<sub>n+1</sub> is very easy, you just have to locate both ends of S<sub>n+1</sub> amongs the ends of D<sub>1</sub>...D<sub>k</sub>.</li> <li>assuming that you know the union of segments S<sub>1</sub>...S<sub>n</sub> consists of disjoints segments D<sub>1</sub>...D<sub>k</sub>, removing segment S<sub>i</sub> (assuming that S<sub>i</sub> was included in D<sub>j</sub>) means recomputing the union of segments that D<sub>j</sub> consisted of, except S<sub>i</sub> (using the static algorithm).</li> </ul> <p>This is your dynamic algorithm. Assuming that you will use sorted sets with log-time location queries to represent D<sub>1</sub>...D<sub>k</sub>, this is probably the most efficient non-specialized method you can get.</p>
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    1. COYes, but after removing a segment S from the set you have to do the *recomputing* which is O(n) in the worst case. So the total running time O(n^2), isn't it? Imagine this case: http://bit.ly/8Zd5jO
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      1. USMartin Konicek
    2. CO@Martin: Unless you use a data structure like an interval tree. That would give you `O(nlogn)`.
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      1. USThomas Ahle
 

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