Bomberman lives in a rectangular grid. Each cell in the grid either contains a bomb or nothing at all. Each bomb can be planted in any cell of the grid but once planted, it will detonate after exactly 3 seconds. Once a bomb detonates, it's destroyed — along with anything in its four neighboring cells. This means that if a bomb detonates in cell , any valid cells and are cleared. If there is a bomb in a neighboring cell, the neighboring bomb is destroyed without detonating, so there's no chain reaction. Bomberman is immune to bombs, so he can move freely throughout the grid. Here's what he does: Initially, Bomberman arbitrarily plants bombs in some of the cells, the initial state. After one second, Bomberman does nothing. After one more second, Bomberman plants bombs in all cells without bombs, thus filling the whole grid with bombs. No bombs detonate at this point. After one more second, any bombs planted exactly three seconds ago will detonate. Here, Bomberman stands back and observes. Bomberman then repeats steps 3 and 4 indefinitely. Note that during every second Bomberman plants bombs, the bombs are planted simultaneously (i.e., at the exact same moment), and any bombs planted at the same time will detonate at the same time. Given the initial configuration of the grid with the locations of Bomberman's first batch of planted bombs, determine the state of the grid after seconds. For example, if the initial grid looks like: ....O.... it looks the same after the first second. After the second second, Bomberman has placed all his charges: OOOOOOOOO At the third second, the bomb in the middle blows up, emptying all surrounding cells: O.O...O.O Function Description Complete the bomberMan function in the editory below. bomberMan has the following parameter(s): int n: the number of seconds to simulate string grid[r]: an array of strings that represents the grid Returns string[r]: n array of strings that represent the grid in its final state Input Format The first line contains three space-separated integers , , and , The number of rows, columns and seconds to simulate. Each of the next lines contains a row of the matrix as a single string of characters. The . character denotes an empty cell, and the O character (ascii 79) denotes a bomb. Constraints Subtask for of the maximum score. Sample Input STDIN Function ----- -------- 6 7 3 r = 6, c = 7, n = 3 ....... grid =['.......', '...O...', '....O..',\ ...O... '.......', 'OO.....', 'OO.....'] ....O.. ....... OO..... OO..... Sample Output OOO.OOO OO...OO OOO...O ..OO.OO ...OOOO ...OOOO Explanation The initial state of the grid is: ....... ...O... ....O.. ....... OO..... OO..... Bomberman spends the first second doing nothing, so this is the state after 1 second: ....... ...O... ....O.. ....... OO..... OO..... Bomberman plants bombs in all the empty cells during his second second, so this is the state after 2 seconds: OOOOOOO OOOOOOO OOOOOOO OOOOOOO OOOOOOO OOOOOOO In his third second, Bomberman sits back and watches all the bombs he planted 3 seconds ago detonate. This is the final state after seconds: OOO.OOO OO...OO OOO...O ..OO.OO ...OOOO ...OOOO
Bomberman lives in a rectangular grid. Each cell in the grid either contains a bomb or nothing at all.
Each bomb can be planted in any cell of the grid but once planted, it will detonate after exactly 3 seconds. Once a bomb detonates, it's destroyed — along with anything in its four neighboring cells. This means that if a bomb detonates in cell , any valid cells and are cleared. If there is a bomb in a neighboring cell, the neighboring bomb is destroyed without detonating, so there's no chain reaction.
Bomberman is immune to bombs, so he can move freely throughout the grid. Here's what he does:
- Initially, Bomberman arbitrarily plants bombs in some of the cells, the initial state.
- After one second, Bomberman does nothing.
- After one more second, Bomberman plants bombs in all cells without bombs, thus filling the whole grid with bombs. No bombs detonate at this point.
- After one more second, any bombs planted exactly three seconds ago will detonate. Here, Bomberman stands back and observes.
- Bomberman then repeats steps 3 and 4 indefinitely.
Note that during every second Bomberman plants bombs, the bombs are planted simultaneously (i.e., at the exact same moment), and any bombs planted at the same time will detonate at the same time.
Given the initial configuration of the grid with the locations of Bomberman's first batch of planted bombs, determine the state of the grid after seconds.
For example, if the initial grid looks like:
it looks the same after the first second. After the second second, Bomberman has placed all his charges:
At the third second, the bomb in the middle blows up, emptying all surrounding cells:
Function Description
Complete the bomberMan function in the editory below.
bomberMan has the following parameter(s):
- int n: the number of seconds to simulate
- string grid[r]: an array of strings that represents the grid
Returns
- string[r]: n array of strings that represent the grid in its final state
Input Format
The first line contains three space-separated integers , , and , The number of rows, columns and seconds to simulate.
Each of the next lines contains a row of the matrix as a single string of characters. The . character denotes an empty cell, and the O character (ascii 79) denotes a bomb.
Constraints
Subtask
- for of the maximum score.
Sample Input
Sample Output
Explanation
The initial state of the grid is:
....... ...O... ....O.. ....... OO..... OO.....Bomberman spends the first second doing nothing, so this is the state after 1 second:
....... ...O... ....O.. ....... OO..... OO.....Bomberman plants bombs in all the empty cells during his second second, so this is the state after 2 seconds:
OOOOOOO OOOOOOO OOOOOOO OOOOOOO OOOOOOO OOOOOOOIn his third second, Bomberman sits back and watches all the bombs he planted 3 seconds ago detonate. This is the final state after seconds:
OOO.OOO OO...OO OOO...O ..OO.OO ...OOOO ...OOOOThis section considers the normalizer of a subgroup H of G. We assume we have a base and
strong generating set for H. The property is
PH : g ~ G normalizes H <G
The two facts we know about normalizing elements which lead to our choice of a base and the
restrictions on the base images are
Lemma 1
If g normalizes a subgroup H then g permutes the orbits of H.
Lemma 2
If g normalizes a subgroup H and g fixes [3 then g normalizes the stabiliser H[~.
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