HDU3999-The order of a Tree

The order of a Tree

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 64 Accepted Submission(s) : 34

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:

1. insert a key k to a empty tree, then the tree become a tree with

only one node;

2. insert a key k to a nonempty tree, if k is less than the root ,insert

it to the left sub-tree;else insert k to the right sub-tree.

We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.

Input

There are multiple test cases in an input file. The first line of each testcase is an integer n(n <= 100,000),represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence
of 1 to n.

Output

One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

Sample Input

4

1 3 4 2

Sample Output

1 3 2 4

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

#define N 100005
int tree
,l
,r
,a
,num,flag,m;

void fx(int index,int x)
{
if( x <= tree[index] )
{
if(l[index] == -1) l[index] = flag;
else fx(l[index],x);
}
else
{
if(r[index] == -1) r[index] = flag;
else fx(r[index],x);
}
}

void f(int index)
{
a[m++] = tree[index];
if(l[index] != -1)
f(l[index]);
if(r[index] != -1)
f(r[index]);
}
int main()
{
int n,i,x,root;
while(~scanf("%d",&n))
{
memset(l,-1,sizeof(l));
memset(r,-1,sizeof(r));
root = -1;
flag = 0;
for(i = 0; i< n; i++)
{
scanf("%d",&x);
if(root == -1)
tree[++root] = x;
else
{
tree[++flag] = x;
fx(root,x);
}
}
m = 0;
f(root);
for(i = 0; i < m -1; i++)
printf("%d ",a[i]);
printf("%d\n",a[m-1]);
}
return 0;
}