  # Résolution de problèmes algorithmiques

Maintain a numerical table that implements the following operations in logarithmic time: query the entry at some index, add a value to all entries between two given indices.

## Binary tree

Without loss of generality suppose that the table has a size n in the form of a power of 2. The idea is to work with a binary tree with n leafs, each corresponding to an entry in the table. Every node p in the tree has a value label p.val, including the leafs. The relation between the tree and the table is as follows. The total value along the path from the root to a leaf equals the corresponding table entry. In that sense the tree encodes the table. The data structure consists only of the tree, not of the table. As we have seen querying the entry at some index can be done in logarithmic time (which is the depth of the tree). Now we explain how to perform table updates. To each node p we associate the interval formed by all the table indices corresponding to the leafs of the subtree. We call this interval p.span. For a leaf p, the interval p.span consists only of the corresponding table index. For an inner node p with descendants p.left and p.right, the interval p.span is the (disjoint) union of p.left.span and p.right.span.

Updates are done by descending recursively the tree, as follows. Note that the span is not stored with the nodes but implicitly given as a parameter with each call.

To convince yourself that the update procedure, when initiated at the root, terminates in logarithmic time, simply observe that for each level k there will be at most two calls to update with p_span of size $2^k$. ## Extension

A really nice feature of segment trees is that they allow you to maintain additional information about the table, which can be quickly queried. Suppose you want to query the sum of the table entries between two indices. To answer these queries quickly we store in each node p a score, which is the sum of the table entries over all indices in p.span. Hence

p.score = p.val + p.left.score + p.right.score


The to answer the query we have to make recursive calls as follows.

## More extensions

Similarly it is possible to adapt segment trees to answer min or max queries. For this purpose one only need to specify

• a neutral value to be returned on queries over empty intervals. This can be 0 for sum-queries, $+\infty$ for min-queries, or $-\infty$ for max-queries.
• a score-update procedure to be applied on leafs, when their value has changed
• a score-udpate procedure to be applied on inner nodes, when their value has changed
• a combine-query function that computes the answer to a query from the answers obtained from each recursive query on the node successors.