HDU 2196 computer (tree DP)

Given a tree, find the maximum path length that each node can reach.

Analysis: Taking the numbered I node as an example (non-root node), the longest path length is only two possibilities, 1) the longest path exists in the subtree; 2) the longest path exists in the path of the parent node

Therefore, you only need to find the two paths corresponding to each node to obtain the maximum value. Of course, the root node only has the first possibility.

View code

# Include <iostream>
# Include <algorithm>
# Include < String >
# Include <vector>
 Using   Namespace STD;
 Const   Int N = 10010 ;
Int N, DIST [N], DP [N];
 Struct Edge
{
 Int V, W, h;
Edge (){}
Edge ( Int A, Int B, Int C = 0 ): V (a), w (B), H (c ){}
};
 Struct Node
{
 Int W, ID;
Node (){}
Node ( Int A, Int B): w (A), ID (B ){}
};
Vector <edge> G [N];
 Void Init ()
{
 For ( Int I = 0 ; I <= N; I ++)
G [I]. Clear ();
}
 Void Dfs1 ( Int U)
{
Int Size = G [u]. Size ();
Dist [u] = 0 ;
Node mdis1 = node ( 0 , 0 ), Mdis2 = node ( 0 , 0 );
 //  Mdis1 and mdis2 respectively record the longest path length in the U subtree and the branch number  
      For ( Int I = 0 ; I <size; I ++)
{
 Int V = G [u] [I]. V;
Dfs1 (v );
Node T = node (G [u] [I]. W + dist [v], I );
 If (T. W> mdis1.w)
{
Mdis2 = mdis1;
Mdis1 = T;
}
 Else   If (T. W> mdis2.w)
Mdis2 = T;
}
Dist [u] = mdis1.w;
 For ( Int I = 0 ; I <size; I ++)
{
 If (I = mdis1.id)
G [u] [I]. H = mdis2.w; //  Record the longest path except the branch where the son of I is located in the subtree.  
          Else G [u] [I]. H = mdis1.w;
}
}
 Void Dfs2 ( Int U)
{
 Int Size = G [u]. Size ();
For ( Int I = 0 ; I <size; I ++)
{
 Int V = G [u] [I]. V;
 If (U = 1 )
{
DP [v] = G [u] [I]. H + G [u] [I]. W;
}
 Else {
DP [v] = max (DP [u], G [u] [I]. h) + G [u] [I]. W; //  The maximum length of the path stored by Father's Day 
 }
Dfs2 (v );
}
}
 Int Main ()
{
 Int X, W;
 While (Scanf ( "  % D  " , & N) = 1 )
{
Init ();
 For ( Int I = 2 ; I <= N; I ++)
{
Scanf ( "  % D  " , & X, & W );
G [X]. push_back (edge (I, W, 0 ));
}
Dfs1 ( 1 );
Memset (DP, 0 , Sizeof (DP ));
DP [ 1 ] = DIST [1 ];
Dfs2 ( 1 );
 For ( Int I = 1 ; I <= N; I ++)
Printf ( "  % D \ n  " , Max (DP [I], DIST [I]);
}
 Return   0 ;
}