2015-Ali C + + development additional questions first question

Similar to the implementation of a Cartesian tree:

Http://baike.baidu.com/link?url=nDGlKta6rvBzJ4_xm1TSp-Px05bU6gzgqWx7LnWwnXm5brtOCPJXXXRXeq9ZpIvsfrWkhua4D76oWrt2iQq2WK

From the topic can be seen, according to a, is a binary sort of tree, according to B, is a big pile.

Then you need to know a property: A two-fork sorting tree in the middle sequence traversal, you can get an ascending sequence.

(1) Build:

Solution 1: pair[] In accordance with the size B to the small sort, so that the first, must be the root node of the tree, continue to traverse pair[], according to the value of a to allocate the location of this node.

For example T->a > insert.a, so that the T right subtree to continue to find, if T->a < insert.a, go to T right subtree continue to find, each time t from the root node.

This efficiency is Nlog (N);

Solution 2: pair[] According to a small to large sort, and then traverse pair[] to build the tree, because the order is over, so with insert inserts must be in the current tree the largest one,

This can happen in two cases: 1. insert.b > t.b, then insert replaces the position of T, then insert.left = t;

2.insert.b < t.b, then, continue to compare with t.right.b, because, INSERT.A is the current largest, definitely go to the right, until a 1 situation occurs, or to a single empty tree, then insert.

This efficiency is also for Nlog (N);

Solution 3: Improved on the basis of solution 2, because each time it is from the root node to compare, it will take a little more time, and then we will use a stack to save the root node, the root node of the right sub-tree nodes (bottom of the root node) .... Since the inserted a is the largest, it must first be compared to the right, 1. If insert.b < stack.top.b, then Stack.top.right = INSERT, and then a right node, into the stack; 2. If insert.b > stack.t OP.B, it is necessary to backtrack, and then the stack--. top.b comparison, if there are 1 of the situation, then, before that right subtree, become insert.left. Insert into the stack, or appear, all stacks go out, indicating that the insert is about to become a new root. Using space to change time, efficiency is O (N);

(2) Insert ();

Solution: First according to the nature of the binary sorting tree, according to the size of a is inserted into a leaf node, after insertion, b judgment, if it is Zuozi inserted, and the inserted B is larger, then the parent node to a right rotation, conversely, if the right subtree is inserted, and the inserted B is larger, A left rotation is performed with his parent node, and back to the root junction.

　　　

voidTreap_insert (Treap_node*&a,intLabelintp) {if(!a) {a=NewTreap_node; A->label=label; A->p=p; }        Else if(label<a->label) {Treap_insert (a-left,label,p); if(a->left->p>a->p) treap_right_rotate (a); }        Else{Treap_insert (a-right,label,p); if(a->right->p>a->p) treap_left_rotate (a); }    }

2015-Ali C + + development additional questions first question