Scissors and Tape
Contest: CEOI 2019 day 2 online mirror (unrated, IOI format)
Tags: *special, constructive algorithms, geometry

You are given a piece of paper in the shape of a simple polygon $$$S$$$. Your task is to turn it into a simple polygon $$$T$$$ that has the same area as $$$S$$$.

You can use two tools: scissors and tape. Scissors can be used to cut any polygon into smaller polygonal pieces. Tape can be used to combine smaller pieces into larger polygons. You can use each tool multiple times, in any order.

The polygons given in the input have integer coordinates, but you are allowed to produce shapes with non-integer coordinates in your output.

A formal definition of the task follows.

A shape $$$Q=(Q_0,\dots,Q_{n-1})$$$ is a sequence of three or more points in the plane such that:

- The closed polyline $$$Q_0Q_1Q_2\dots Q_{n-1}Q_0$$$ never touches or intersects itself, and therefore it forms the boundary of a simple polygon.
- The polyline goes around the boundary of the polygon in the counter-clockwise direction.

The polygon whose boundary is the shape $$$Q$$$ will be denoted $$$P(Q)$$$.

Two shapes are called equivalent if one can be translated and/or rotated to become identical with the other.

Note that mirroring a shape is not allowed. Also note that the order of points matters: the shape $$$(Q_1,\dots,Q_{n-1},Q_0)$$$ is not necessarily equivalent to the shape $$$(Q_0,\dots,Q_{n-1})$$$.

In the figure on the left: Shapes $$$U$$$ and $$$V$$$ are equivalent. Shape $$$W$$$ is not equivalent with them because the points of $$$W$$$ are given in a different order. Regardless of the order of points, the fourth shape is not equivalent with the previous ones either as flipping a shape is not allowed.

In both input and output, a shape with $$$n$$$ points is represented as a single line that contains $$$2n+1$$$ space-separated numbers: the number $$$n$$$ followed by the coordinates of the points: $$$Q_{0,x}$$$, $$$Q_{0,y}$$$, $$$Q_{1,x}$$$, ...

Shapes have identification numbers (IDs). The given shape $$$S$$$ has ID 0, the shapes you produce in your solutions are given IDs 1, 2, 3, ..., in the order in which they are produced.

Shapes $$$B_1,\dots,B_k$$$ form a subdivision of shape $$$A$$$ if:

- The union of all $$$P(B_i)$$$ is exactly $$$P(A)$$$.
- For each $$$i\neq j$$$, the area of the intersection of $$$P(B_i)$$$ and $$$P(B_j)$$$ is zero.

The scissors operation destroys one existing shape $$$A$$$ and produces one or more shapes $$$B_1,\dots,B_k$$$ that form a subdivision of $$$A$$$.

In the figure: Shape $$$A$$$ (square) subdivided into shapes $$$B_1$$$, $$$B_2$$$, $$$B_3$$$ (the three triangles). One valid way to describe one of the $$$B_i$$$ is "3 3 1 6 1 5.1 4".

The tape operation destroys one or more existing shapes $$$A_1,\dots,A_k$$$ and produces one new shape $$$B$$$. In order to perform this operation, you must first specify shapes $$$C_1,\dots,C_k$$$ and only then the final shape $$$B$$$. These shapes must satisfy the following:

- For each $$$i$$$, the shape $$$C_i$$$ is equivalent to the shape $$$A_i$$$.
- The shapes $$$C_1,\dots,C_k$$$ form a subdivision of the shape $$$B$$$.

Informally, you choose the shape $$$B$$$ and show how to move each of the existing $$$A_i$$$ to its correct location $$$C_i$$$ within $$$B$$$. Note that only the shape $$$B$$$ gets a new ID, the shapes $$$C_i$$$ do not.

Input format:
The first line contains the source shape $$$S$$$.

The second line contains the target shape $$$T$$$.

Each shape has between 3 and 10 points, inclusive. Both shapes are given in the format specified above.

All coordinates in the input are integers between $$$-10^6$$$ and $$$10^6$$$, inclusive.

In each shape, no three points form an angle smaller than 3 degrees. (This includes non-consecutive points and implies that no three points are collinear.)

The polygons $$$P(S)$$$ and $$$P(T)$$$ have the same area.

Output format:
Whenever you use the scissors operation, output a block of lines of the form:

where $$$id(A)$$$ is the ID of the shape you want to destroy, $$$k$$$ is the number of new shapes you want to produce, and $$$B_1,\dots,B_k$$$ are those shapes.

Whenever you use the tape operation, output a block of lines of the form:

where $$$k$$$ is the number of shapes you want to tape together, $$$id(A_1),\dots,id(A_k)$$$ are their IDs, $$$C_1,\dots,C_k$$$ are equivalent shapes showing their position within $$$B$$$, and $$$B$$$ is the final shape obtained by taping them together.

It is recommended to output coordinates of points to at least 10 decimal places.

The output must satisfy the following:

- All coordinates of points in the output must be between $$$-10^7$$$ and $$$10^7$$$, inclusive.
- Each shape in the output must have at most $$$100$$$ points.
- In each operation the number $$$k$$$ of shapes must be between $$$1$$$ and $$$100$$$, inclusive.
- The number of operations must not exceed $$$2000$$$.
- The total number of points in all shapes in the output must not exceed $$$20000$$$.
- In the end, there must be exactly one shape (that hasn't been destroyed), and that shape must be equivalent to $$$T$$$.
- All operations must be valid according to the checker. Solutions with small rounding errors will be accepted. (Internally, all comparisons check for absolute or relative error up to $$$10^{-3}$$$ when verifying each condition.)

Constraints:
Time limit: 1.0 s
Memory limit: 1024.0 MB

Note:
Scoring:
A shape is called a nice rectangle if it has the form $$$((0,0),~ (x,0),~ (x,y),~ (0,y))$$$ for some positive integers $$$x$$$ and $$$y$$$.

A shape is called a nice square if additionally $$$x=y$$$.

A shape $$$A$$$ is called strictly convex if all inner angles of the polygon $$$P(A)$$$ are smaller than 180 degrees.

Subtask 1 (5 points): $$$S$$$ and $$$T$$$ are nice rectangles. All coordinates of all points are integers between 0 and 10, inclusive

Subtask 2 (13 points): $$$S$$$ is a nice rectangle with $$$x>y$$$, and $$$T$$$ is a nice square

Subtask 3 (12 points): $$$S$$$ and $$$T$$$ are nice rectangles

Subtask 4 (14 points): $$$S$$$ is a triangle and $$$T$$$ is a nice square

Subtask 5 (10 points): $$$S$$$ and $$$T$$$ are triangles

Subtask 6 (16 points): $$$S$$$ is a strictly convex polygon and $$$T$$$ is a nice rectangle

Subtask 7 (11 points): $$$T$$$ is a nice rectangle

Subtask 8 (19 points): no additional constraints

The figure below shows the first example output. On the left is the original figure after using the scissors, on the right are the corresponding $$$C_i$$$ when we tape those pieces back together.

In the second example output, note that it is sufficient if the final shape is equivalent to the target one, they do not have to be identical.

The figure below shows three stages of the third example output. First, we cut the input rectangle into two smaller rectangles, then we cut the bigger of those two rectangles into two more. State after these cuts is shown in the top left part of the figure.

Continuing, we tape the two new rectangles together to form a six-sided polygon, and then we cut that polygon into three 2-by-1 rectangles and one smaller rectangle. This is shown in the bottom left part of the figure.

Finally, we take the rectangle we still have from the first step and the four new rectangles and we assemble them into the desired 3-by-3 square.

Example input:
6 0 0 6 0 6 4 5 4 5 9 0 9
4 0 0 7 0 7 7 0 7

Example output:
scissors
0 5
3 0 0 3 0 3 4
3 3 4 0 4 0 0
3 3 0 6 0 6 4
3 6 4 3 4 3 0
4 0 4 5 4 5 9 0 9
tape
5 1 2 5 3 4
3 0 3 0 0 4 0
3 4 0 7 0 7 4
4 0 3 4 0 7 4 3 7
3 7 4 7 7 3 7
3 3 7 0 7 0 3
4 0 0 7 0 7 7 0 7

Example input:
4 0 0 3 0 3 3 0 3
4 7 -1 10 -1 11 2 8 2

Example output:
scissors
0 2
3 0 0 1 3 0 3
4 1 3 0 0 3 0 3 3
tape
2 1 2
3 110 -1 111 2 110 2
4 108 2 107 -1 110 -1 110 2
4 107 -1 110 -1 111 2 108 2

Editorial:
There are many correct approaches. We will describe one that's reasonably simple to implement. We'll start with some basic shapes and we'll work our way up to general polygons.
Step one: a nice rectangle to another nice rectangle
We will do this in a series of steps. While the input rectangle is at least twice as wide (or twice as tall) as the other, cut it in half and place those halves one on top of the other. For the final step (if needed) we can go directly from one rectangle to the other by cutting it into three pieces and shifting two of them, as shown in the figure below.

Step two: any triangle to a nice rectangle
Move the input triangle so that its longest side is on the x-axis and its other vertex has positive y. (You can "abuse" the tape operation with $$$k=1$$$ to do this.) Then, cut it horizontally at half of its height and cut the top part vertically through the vertex. The three pieces can be reassembled to form a rectangle.
Step three: a nice rectangle to a triangle
This is just step two backwards: resize the rectangle to be as wide as the longest side of the triangle, then cut it in three pieces and assemble the triangle you need.

For the full construction, we can now go as follows: Triangulate the input polygon. Change each triangle into a rectangle. Resize all rectangles to have the same height. Tape them to a single rectangle $$$R$$$ of that height.
Triangulate the output polygon to know the areas of its triangles. Cut $$$R$$$ into rectangles of those areas. Transform each rectangle to the corresponding triangle, and tape them all together.
An alternate full construction is can be implemented by having "triangle to triangle" as a primitive and then taking the triangulations of $$$S$$$ and $$$T$$$ and additionally cutting some of their triangles into two so that you'll get two collections of triangles of matching areas. (Imagine having a list of triangles from $$$S$$$ and a list of triangles from $$$T$$$. You read both lists at the same time and whenever the areas don't match you cut off and use an appropriate part of the bigger triangle.)

Historic background: The equivalence relation between objects that can be obtained by a finite number of our "scissors" and "tape" steps is better known under the name scissors congruence. Wallace, Bolyai and Gerwien were the first to show that in 2D any two polygons with the same area are scissors-congruent. The 3D version of this problem was one of Hilbert's problems, and it was the first one solved – with a negative result. Within a year of Hilbert's talk, his student Dehn constructed two polyhedra with the same volume that are not scissors-congruent.