Conclusions about tree arrays from multi-School Joint nudt Question 2 cube

At a glance, this question is a typical tree array solution for updating the interval lookup points. However, if a three-dimensional tree array is used, there will be a problem: how to split it?

Let's start with a simple question. For one-dimensional situations, we can update the intervals (a, B), so we can add v At A, B + 1 minus V.

For two-dimensional situations, it is troublesome to update the region (x1, Y1) to (X2, Y2), then add V at (x1, Y1, subtract V at (X2 + 1, Y1) and (x1, y2 + 1), and add V at (X2 + 1, y2 + 1.

So what about 3D? After drawing, I got the conclusion that (x1, Y1, Z1) plus V, and (x1, Y1, Z1) minus V, these points include (X2 + 1, Y1, Z1), (x1, y2 + 1, Z1), (x1, Y1, Z2 + 1), and (X2 + 1, y2 + 1, Z2 + 1), and the rest is v.

The three tree arrays with different dimensions are combined. The source vertex P adds V and P minus V at the adjacent odd position, and the even position is added v, in this way, you can quickly write down all the cutting conditions.

 

My code:

 

# Include <iostream> <br/> # include <cstdio> <br/> # include <cstring> <br/> # include <cstdlib> </P> <p> using namespace STD; </P> <p> const int max = 110; <br/> int N; <br/> int tree [Max] [Max] [Max]; </P> <p> void add (int x, int y, int Z, int v) <br/> {<br/> int tmp1 = y; <br/> int tmp2 = z; <br/> while (x <= N) <br/> {<br/> Y = tmp1; <br/> while (Y <= N) <br/> {<br/> Z = tmp2; <br/> while (z <= N) <br/> {<br/> tree [x] [y] [Z] + = V; <br/> Z + = Z &-Z; <BR/>}< br/> Y + = Y &-y; <br/>}< br/> X + = x &-X; <br/>}</P> <p> int read (int x, int y, int Z) <br/>{< br/> int tmp1 = y; <br/> int tmp2 = z; <br/> int sum = 0; <br/> while (X) <br/>{< br/> Y = tmp1; <br/> while (y) <br/>{< br/> Z = tmp2; <br/> while (z) <br/> {<br/> sum + = tree [x] [y] [Z]; <br/> Z-= Z &-Z; <br/>}< br/> Y-= Y &-y; <br/>}< br/> X-= x &-X; <br/>}</P> <p> return sum; <br/>}</P> <p> int main () <br/>{< br/> int m; <br/> int x1, x2, Y1, Y2, Z1, Z2, OP; </P> <p> while (scanf ("% d", & N, & M )! = EOF) <br/>{< br/> for (INT I = 1; I <= N; I ++) <br/> for (Int J = 1; j <= N; j ++) <br/> for (int K = 1; k <= N; k ++) <br/> tree [I] [J] [k] = 0; <br/> for (INT I = 0; I <m; I ++) <br/>{< br/> scanf ("% d", & OP); <br/> If (OP) <br/>{< br/> scanf ("% d", & X1, & Y1, & Z1 ); <br/> scanf ("% d", & X2, & Y2, & Z2); <br/> Add (x1, Y1, Z1, 1 ); <br/> Add (X2 + 1, Y1, Z1,-1); <br/> Add (x1, y2 + 1, Z1,-1 ); <br/> Add (x1, Y1, Z2 + 1,-1); <br/> Add (X2 + 1, y2 + 1, Z1, 1 ); <br/> Add (X2 + 1, Y1, Z2 +); <br/> Add (x1, y2 + 1, Z2 + ); <br/> Add (X2 + 1, y2 + 1, Z2 + 1,-1 ); <br/>}< br/> else <br/> {<br/> scanf ("% d", & X1, & Y1, & Z1 ); <br/> printf ("% d/N", read (x1, Y1, Z1) % 2 ); <br/>}</P> <p> return 0; <br/>}< br/>

 

Conclusion: after each question is done, you need to reflect on it. If you don't have to do one question, you 'd better spend time optimizing the code. After all, this is better for you. The question is not expensive, but expensive.