NOIP 2004 universal group rematch FBI Tree

NOIP 2004 universal group rematch FBI Tree

1. Read the topic, but also some unintelligible.

2. Manual simulation of the sample to understand the test instructions.

10001011

The FBI tree is shown in the figure below

F

F F

F B F I

I b b b I b i I

1) the root node of T is R, whose type is the same as the type of string s, and this sentence is the core of the core, that is, the F B I three kinds of root nodes.

3. The next programming implementation is the focus. First, the children believe that a layer of achievement, with a one-dimensional array storage, the 1th (No. No. 0) element is used to start using the array. Father K, left child 2*k, right son 2*k+1. Then the traversal, the post-sequential traversal, the use of recursive way.

4. Build function, traverse function to write.

5. Write the code, submit the AC, recursive function, and remember to write the termination condition first.

6. This article is well written, readers can learn from it. http://blog.csdn.net/dogeding/article/details/52727239

7. This problem can be used for reference: Two tree storage, binary tree traversal.

Attach the AC code, compile the Environment dev-c++4.9.9.2

#include <stdio.h>

int n;
Char a[1024+10];
Char b[2048+10];

void Build (char *s,int N) {//Achievement code
Underlying structure
int i,j=0;
For (i= (1<<n); i<= (1<< (n+1)) -1;i++) {
b[i]=a[j++]== ' 0 '? B ': ' I ';
}

Bottom-up construction
for (i=n-1;i>=0;i--)
For (j= (1<<i); j<= (1<< (i+1)) -1;j++)
if (b[2*j]== ' B ' &&b[2*j+1]== ' B ')
b[j]= ' B ';
else if (b[2*j]== ' i ' &&b[2*j+1]== ' I ')
b[j]= ' I ';
Else
b[j]= ' F ';
}
void visit (int node) {//post-Traversal code
if (node> (1<< (n+1))-1)
Return
Visit (node*2);//Zuozi
Visit (node*2+1);//Right Sub-tree
printf ("%c", B[node]);//Root
}
int main () {
scanf ("%d", &n);
scanf ("%s", a);
Build (A,n);
Visit (1);
printf ("\ n");
return 0;
}