UVA 10410 treereconstruction

Test instructions, give you a tree of the BFS sequence and DFS sequence, the node number of a small priority calendar, ask you a possible tree shape;

Outputs the sub-nodes of each node.

Note the following facts:

(1) The sub-tree nodes of a node in the DFS sequence must be contiguous.

(2) The subsequent node of a node u in the Bfs,dfs sequence must be either U or U's post-sibling junction {v}, or the descendant nodes {s} of U and {v}.

(3) If there is a post-sibling junction, then the BFS sequence in the back of U is bound to be the first after the brothers knot Point V1,

(4) If there are descendant nodes, then the first child node must be followed by U in the DFS sequence S1.

The BFS sequence of the Memory node U is the BFS (U) and the DFS sequence is DFS (v).

In the DFS sequence, a junction of U, node v satisfies dfs (v) = DFS (u) + 1 if BFS (v) = BFS (u) +1 and V > U; then v must be regarded as the first posterior sibling node of u,

If not, then V is the child node of U, you can launch U is the last node of the U layer in the BFS, when you do not have a post-sibling node, so the next node must be his descendants node, then v must be equivalent to the sibling node of U without changing the bfs,dfs sequence.

To this, (5) to meet the BFS (v) = BFS (u) +1 and V > U condition of V as the first post-sibling node of U, do not meet this condition must not be the post-sibling node, this can be based on the definition can be proven.

If v satisfies (5), according to (1), you and the subtree are accessed, if V does not satisfy the condition and BFS (v) >bfs (u) + 1 then v must be a sub-node of U, if BFS (v) <bfs (U) then indicates that V is its parent node, and that the subtree of U has been visited.

Iterate over the above process, the boundary condition is root. It's done!

//Rey#include <bits/stdc++.h>using namespacestd;Const intMAXN = ++5; Vector<int>G[MAXN];intPOS[MAXN];intMain () {//freopen ("In.txt", "R", stdin);   intN; intT;  while(~SCANF ("%d", &n) &&N) {       for(inti =1; I <= N; i++) scanf ("%d", &t), pos[t] =I, g[i].clear (); intRoot; scanf ("%d",&root); Stack<int>STA;      Sta.push (root);  for(inti =1; I < n; i++) {scanf ("%d",&t);  for(;;) {            intU =Sta.top (); if(pos[u]+1< Pos[t] | | (pos[u]+1= = Pos[t] && u > t) | | U = =root)               {g[u].push_back (t);               Sta.push (t);  Break; }Else{sta.pop (); }         }      }       for(inti =1; I <= N; i++) {printf ("%d:", i);  for(intj =0, Sz = G[i].size (); J < Sz; J + +) printf ("%d", G[i][j]); Puts (""); }   }   return 0;}

PS: Blogging is really good to exercise their ability to express, but also to comb the previous ideas.

UVA 10410 treereconstruction