Poj 3246 Balanced Lineup (line segment tree basis)

The topic Description describes a binary tree that outputs its maximum width and height. The Input Description is an integer n in the first line of Input Description. There are two numbers in each row in the n rows below, which is the number of the node connected to the binary tree. If the n rows are not connected to the node, the value is 0. Output Description outputs a total of rows. the maximum width and height of the Output binary tree are separated by a space. Sample input Sample Input5 2 3 4 5 0 0 0 0 0 0 Sample output Sample Output2 3

#include<iostream>   #include<cstdio>   #include<cstring>   #include<cmath>   #include<algorithm>   #include<bitset>   #include<iomanip>     using namespace std;    int a[ 40 ][ 3 ] = { 0 } , num[ 40 ] ;    int w = 0 , d = 0 ;  void dfs( int x , int y )  {      num[ y ]++ ;      if( y > w )          w = y ;      if( a[ x ][ 1 ] != 0 )          dfs( a[ x ][ 1 ] , y + 1 ) ;      if( a[ x ][ 2 ] != 0 )          dfs( a[ x ][ 2 ] , y + 1 ) ;  }  int main()  {      int n ;      cin >> n ;      for( int i = 1 ; i <= n ; i++)          cin >> a[ i ][ 1 ] >> a[ i ][ 2 ] ;      dfs( 1 , 1 ) ;      for( int i = 1 ; i <= 16 ; ++i )          if( num[ i ] > d )              d = num[ i ] ;      cout << d << " " << w << endl ;      return 0 ;  }  #include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<bitset>#include<iomanip>using namespace std;int a[ 40 ][ 3 ] = { 0 } , num[ 40 ] ;int w = 0 , d = 0 ;void dfs( int x , int y ){num[ y ]++ ;if( y > w )w = y ;if( a[ x ][ 1 ] != 0 )dfs( a[ x ][ 1 ] , y + 1 ) ;if( a[ x ][ 2 ] != 0 )dfs( a[ x ][ 2 ] , y + 1 ) ;}int main(){int n ;cin >> n ;for( int i = 1 ; i <= n ; i++)cin >> a[ i ][ 1 ] >> a[ i ][ 2 ] ;dfs( 1 , 1 ) ;for( int i = 1 ; i <= 16 ; ++i )if( num[ i ] > d )d = num[ i ] ;cout << d << " " << w << endl ;return 0 ;}