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Segment tree implementation

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Antti H S Laaksonen 2017-02-18 19:22:02 +02:00
parent 3fb9b598ae
commit 9da26930f3
1 changed files with 5 additions and 5 deletions
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luku28.tex
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@ -37,8 +37,8 @@ Using this approach, the function becomes as follows:
int sum(int a, int b, int k, int x, int y) { int sum(int a, int b, int k, int x, int y) {
if (b < x || a > y) return 0; if (b < x || a > y) return 0;
if (a <= x && y <= b) return p[k]; if (a <= x && y <= b) return p[k];
int d = (y-x+1)/2; int d = (x+y)/2;
return sum(a,b,2*k,x,x+d-1) + sum(a,b,2*k+1,x+d,y); return sum(a,b,2*k,x,d) + sum(a,b,2*k+1,d+1,y);
} }
\end{lstlisting} \end{lstlisting}
Now we can calculate the sum of Now we can calculate the sum of
@ -63,9 +63,9 @@ the sum can be found in \texttt{p}.
If $[x,y]$ is partially inside $[a,b]$, If $[x,y]$ is partially inside $[a,b]$,
the search continues recursively to the the search continues recursively to the
left and right half of $[x,y]$. left and right half of $[x,y]$.
The size of both halves is $d=\frac{1}{2}(y-x+1)$; The left half is $[x,d]$
the left half is $[x,x+d-1]$ and the right half is $[d+1,y]$
and the right half is $[x+d,y]$. where $d=\lfloor \frac{x+y}{2} \rfloor$.
\end{samepage} \end{samepage}
The following picture shows how the search proceeds The following picture shows how the search proceeds