Type: Posts; User: Zachm
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September 7th, 2011, 03:48 PM
This is in fact a clustering problem - you wish to cluster n airports to gi groups where each group centroid is a gas station coordinate.
Simplest (but maybe not best) method is to use the K-means...
"Lists have the important property that insertion and splicing do not invalidate iterators to list elements, and that even removal invalidates only the iterators that point to the...
If you *KNOW* there are exactly n/2 1's and n/2 2's, then why sort ?
Go over the array and assign 1 to the first n/2 elements and 2 for the rest.
Exactly n assignments and no need for comparison......
Basically you are rearranging each of the matrices rows according to the sorted ordering of a given column in a given matrix.
You just need to figure out what is the new row index of each of the...
April 26th, 2011, 12:21 AM
Mathematica is also a good option, in my experience.
March 16th, 2011, 03:56 PM
Computationally speaking, just emulating the program run and counting the number of steps is not so hard - but if a program with finite run-time may still run for a VERY long time depending on the...
March 13th, 2011, 06:30 AM
Seems to me that a ternary tree could be used the same way as a binary tree is used - by, for example, ignoring the right-most child (keeping it NULL) of each node, and considering the left-most and...
March 12th, 2011, 03:14 PM
As I remember, the number of moves required is (2^n) - 1,
anyhow, this number is Theta(2^n) if and only if it's O(2^n) and also Omega(2^n).
Go back to the mathematical definition of the big-Oh...
March 7th, 2011, 01:04 AM
You can use [ code ] tags to wrap around your code - '#' icon in rich edit mode.
The reason your code goes into an infinite loop is that the for loop exit condition is that i > 255 (since if i <=...
March 7th, 2011, 12:55 AM
Well, that's how cin works.
Good idea! (except for the 'but that seems too complicated' part :)).
If you try doing as you suggested, maybe it won't seem so complicated.
March 6th, 2011, 12:51 AM
To find the ratio between your "world" coordinate units to the cost/heuristic function units, just look at the distance cost between 2 adjacent points relative to the "coordinate" distance between...
February 28th, 2011, 03:20 AM
This is because this is not C++ or C code at all. This is bison(or yacc) BNF input format.
Maybe this yacc calculator example will prove handy ?
I am absolutely *not* the master of bison(or yacc),...
February 9th, 2011, 11:51 PM
February 9th, 2011, 04:49 AM
Radix Sort can give you run-time complexity of O(n).
February 7th, 2011, 01:43 AM
Do you want to find two numbers X, Y, where X is a number from the first array and Y is a number from the second, such that X + Y = S, and S is some predefined value ?
Are the arrays sorted ?
February 7th, 2011, 12:57 AM
Or you can do this:
main(int t,char _,char *a)
return!0<t ? t<3 ? main(-79,-13,a+main(-87,1-_,main(-86, 0, a+1 )+a)):1,t<_?main(t+1, _, a ):3,main ( -94, -27+t, a )&&t==...
February 6th, 2011, 08:15 AM
bool Mob::checkLeader(Mob* L)
if(leader != NULL) //leader exists
if(leader != L) //leader is not me
January 30th, 2011, 07:57 AM
Yes, it is.
Think about how you can sum the value of a path starting at index i, given the sum of the path starting at index i-1, in O(1):
161 = 97 + 32 + 32
164 = 32 + 32 + 100
164 = 161 - 97...
January 28th, 2011, 06:28 AM
This makes it even simpler.
You can think of your segment as having only ranges which are 1 unit in length by dividing each range with weight w of x units length to x ranges with weight w each.
January 23rd, 2011, 12:40 PM
This seems like a rather straight forward problem, though some optimizations come to my mind. There exists an O(n) algorithm to solve this problem, where n is the total number of units.
November 21st, 2010, 02:22 AM
In general, you can use the Flood Fill algorithm in order to find all pixels with the same color, starting from a given pixel.
But what will you do if you have two different circles which are...
November 11th, 2010, 09:40 AM
I have 5 years of commercial development experience in both C++ and C#.
You can PM me if this is still relevant and we will close the details (I couldn't PM you since you...
October 31st, 2010, 03:26 AM
If I put code lines which does the majority of the computation in your example code, it would be the following lines:
cornerSlope = (ymax - p1y) / (xmin - p1x); //in clipEnd()
October 27th, 2010, 02:53 AM
If we're going for an overkill, there are many funny ways of doing stuff ;).
For instance, this some what odd way of approximating Pi :
#define _ F-->00 || F-OO--;
October 24th, 2010, 01:35 PM
I don't think this case is possible since the clique is maximum:
Assume there is some vertex v in the IS which is connected to all vertices of the clique, it follows that degree(v) >= m and...
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