**Problem I. Toys.**

In this problem we need to output all partitions of the given set into subsets in the order which is very similar to the Gray code. Lets denote each partition by a restricted growth string. For a restricted growth string *a* _{1} *a* _{2} *a* _{ n} holds that *a* _{1} = 0 and *a* _{ j + 1} ≤ 1 + *max*(*a* _{1}, ..., *a* _{ j}) for 1 ≤ *j* < *n*. Every partition can be encoded with such string using the following idea: *a* _{ i} = *a* _{ j} if and only if elements *i* and *j* belong to the same subset in the partition. For example, string representation of the partition {1,3},{2},{4} is 0102.

Now we will learn how to generate all restricted growth strings by making a change in exactly one position in the current string to get the next string. It is obvious that in terms of partitions it is what we are asked for in the problem. Rather easy way to build such list of strings was invented by Gideon Ehrlich. Imagine that we have the required list *s* _{1}, *s* _{2}, ..., *s* _{ k} for the length *n* - 1, We will obtain a list for the length *n* from it. Lets *s* _{ i} = *a* _{1} *a* _{2}... *a* _{ n - 1}, and *m* = 1 + *max*(*a* _{1}, ..., *a* _{ n - 1}). Then, if *i* is odd, we will obtain strings of the length *n* by appending digits 0, *m*, *m* - 1, ..., 1 to *s* _{ i}, otherwise we will append digits in order 1, ..., *m* - 1, *m*, 0. Thus, starting from the list 0 for *n* = 1 we will consequently get lists 00, 01 for *n* = 2 and 000, 001, 011, 012, 010 for *n* = 3. Ehrlich scheme is decribed in Knuth's "The art of programming", volume 4, fascicle 3, pages 83-84.

**Problem G. Shooting Gallery.**

Lets solve slightly different problem: for every target we will determine the shoot that hits it. Sort the targets in increasing order of their z-coordinate and process them in that order. Each target is processed as follows. Consider all shoots that potentially can hit it. It is obvious that all such shoots belong to the rectangle, corresponding to the target. From these shoots, the earliest shoot will hit the target. We should find this shoot and remove it from the set of shoots, and then turn to the next target. It's easy to see that the following condition will be held: before we process a target, all shoots that were going to hit it but faced other targer, were already removed from the set of shoots.

Now we need to implement the algorithm efficiently. We will store the shoots in some data structure. This structure should be able to answer two types of queries:

1) Find element with minimum value in the given rectangle.2) Remove the given element.

In my solution I used two-dimensional index tree to manage these queries. I won't describe what the two-dimensional index tree is. I just want to make several remarks. First, the removing operation is not as easy to implement in a two-dimensional index tree as it mays seem. But we are lucky that we have no additions, just deletions! Time complexity of the model solution is *O*((*N* + *M*)*log* ^{2} *N*.

**Problem F. BerPaint.**

Imagine that all segments were drawn. We will refer to these segments as to initial segments. Lets divide the rectangle of drawing into the set of regions and segments such that there are no points of the initial segments strictly inside any region, and new segments separate the regions. Note that new set of segments can contain not only the parts of the initial segments, but also some dummy segments. Initially the color of all regions is white, while the color of each segment can be black of white (dummy segments are white). Please note that in such a partition the border of the region is not consider to belong to it. Lets build a graph where each vertice corresponds either to a region or to a segment, and add edges according to the following rules:

1) Edge between two non-dummy segments is in the graph if these segments have common end-point.2) Edge between a region and a segment (dummy or not) is in the graph if they have more than one common point (i.e. the segment is a part of the border of the region).

It is clear that every region that can be filled corresponds to some connected component of this graph. That gives us a solution. We will store a color for each vertice. When processing a filling operation, we search for all such vertices that the objects that correspond to these vertices contain the chosen point. For region, the point should lie strictly inside the region. For the dummy segment, the point should lie on it but should not coincide with it end-points. And for the non-dummy segment, the point should just lie on it. From each of the found vertices, we make a DFS or BFS which finds all vertices that are reachable from the statring vertice and have the same color, and paints them with new color. After all operations, we need to find sum of areas for such colors, that there are at least one vertice with this color.

The main difficulty in the problem is to divide the rectangle into regions and segments. In my solution it is done using vertical decomposition. First, divide the rectangle into vertical stripes such that inner area of any stripe doesn't contain neiher end-points of the initial segments nor points of their intersections. Then each of these stripes is divided into trapezoid by initial segments, intersecting the stripe. Then add necessary dummy segments to separate the regions and build the graph. I think that there may be some easier ways to construct such graph.

First note that Gray code is 'circular': the last element differs from the first one in one digit. So, we can start enumerating subsets from any position in Gray code.

Now we'll enumerate partitions recursively. Fix the minimal element

xand iterate over the (Gray code of the) subset of elements which are in the same subset withx. IfMis the subset of all other elements, call our procedure recursively forM. We'll obtain a 'good' sequence of partitions. Unfortunately, ifM_{1}andM_{2}are two consequent values ofM, the sequences of partitions for them could not 'glue together': the last partition obtained fromM_{1}can be very different from the first partition obtained fromM_{2}. But we can always apply a circular shift to the second sequence so that it glued together with the first one. I don't know what's the complexity of this but it passed the tests.