Greetings!
I have problems on the "open path" traveling salesman problem.
That is, given a starting point s, an end point d, accompained with n
points, to find the shortest path from s to d and all the n cities
(just once).
Is this a variation of TSP ?
Can anyone advise where can i find relat
main()
{
int arr[n]= {1,4,2,3,5,...};
long SN = n*(n+1)/2;
long SN2 = n*(n+1)*(2*n+1)/6;
long Sn = 0, Sn2= 0;
long D1= 0, D2= 0;
for (int i = 0; i < n ; ++i)
{
Sn += arr[i];
Sn2 += arr[i]*arr[i];
}
D1 = SN - Sn;
D2 = SN2 - Sn2;
X = (D2 + (D1*D1)) / (2*D1);
hi this problem is already solver previously.
check for "Re: Finding Repeated and Missing Numbers"
if X is the missing number and Y is the repeating number.
then let S(N) = n*(n+1)/2
sum of all the numbers provided be Sn
ler S(N2) = n*(n+1)*(2*n+1)/6
sum of square of all the numbers provided be
Although, on second thought, depending on how the graph was laid out,
this might eliminate more weight (value) than the node itself.
You would also have to take into consideration how many nodes (rooms)
are actually connected at any one intersection.
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My suggestion would be to start with an adjacency matrix for the graph.
Take the highest value eliminate adjacent nodes.
Repeat.
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the highest possible value of summed rooms with camera
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carmelo wrote:
> imagine to have a k-ary tree, each node is a room and each edge is a
> corridor; each room has a value assigned, you must place cameras in
> rooms but if a room has an adiacent room with the camera you can't
> place the camera inside...you must place the cameras to have the max
>
imagine to have a k-ary tree, each node is a room and each edge is a
corridor; each room has a value assigned, you must place cameras in
rooms but if a room has an adiacent room with the camera you can't
place the camera inside...you must place the cameras to have the max
value possibile.. how to
