I would like to redefine my algo with cases clarified...
Create a queue that is made to contain the points...
say points queue [1000];
for i:1 to n
for j:i+1 to n
Calculate d (distance between the two centers)
if (d >= r0 + r1) keep them in two separate queues //the circles
don't intersect
if(d==0 || d<= abs(r0-r1))
ignore the circle with smaller radius // one circle
wholly contains another such that the borders do not overlap, or
overlap exactly (e.g. two identical circles)
else
keep both of them in one single queue
Now calculate the area of the circles in those queues which have
single element...
those with more than one element..calculate the area using simple
geometry...You can take help of this..
http://mathworld.wolfram.com/Circle-CircleIntersection.html
Hope its clear now...
On 7/20/11, SAMMM <[email protected]> wrote:
> I doubth .
>
> For (d< r0 + r1) ignore the point with smaller radius as it will
> overshadowed the bigger circle completely
>
> There may be a case where the circle is partially overlapped by the
> other circles. Then this algo will fail .
>
> The area will be of like these :-
>
> Suppose 3 circles are there X,Y&Z .
> Then the area will be :-
>
> Case1:- X+Y+Z
> Case2:- X+(YUZ) ==>> Y + Z - (YnZ) <--- intersection
> case3:- There circle can overlap ... like these .
>
> Then Will your algo work .. I guess no .
>
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