New Theatre Square
Contest: Educational Codeforces Round 88 (Rated for Div. 2)
Difficulty rating: 1000
Tags: brute force, dp, greedy, implementation, two pointers

You might have remembered Theatre square from the problem 1A. Now it's finally getting repaved.

The square still has a rectangular shape of $$$n \times m$$$ meters. However, the picture is about to get more complicated now. Let $$$a_{i,j}$$$ be the $$$j$$$-th square in the $$$i$$$-th row of the pavement.

You are given the picture of the squares:

- if $$$a_{i,j} = $$$ "*", then the $$$j$$$-th square in the $$$i$$$-th row should be black;
- if $$$a_{i,j} = $$$ ".", then the $$$j$$$-th square in the $$$i$$$-th row should be white.

The black squares are paved already. You have to pave the white squares. There are two options for pavement tiles:

- $$$1 \times 1$$$ tiles — each tile costs $$$x$$$ burles and covers exactly $$$1$$$ square;
- $$$1 \times 2$$$ tiles — each tile costs $$$y$$$ burles and covers exactly $$$2$$$ adjacent squares of the same row. Note that you are not allowed to rotate these tiles or cut them into $$$1 \times 1$$$ tiles.

You should cover all the white squares, no two tiles should overlap and no black squares should be covered by tiles.

What is the smallest total price of the tiles needed to cover all the white squares?

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of testcases. Then the description of $$$t$$$ testcases follow.

The first line of each testcase contains four integers $$$n$$$, $$$m$$$, $$$x$$$ and $$$y$$$ ($$$1 \le n \le 100$$$; $$$1 \le m \le 1000$$$; $$$1 \le x, y \le 1000$$$) — the size of the Theatre square, the price of the $$$1 \times 1$$$ tile and the price of the $$$1 \times 2$$$ tile.

Each of the next $$$n$$$ lines contains $$$m$$$ characters. The $$$j$$$-th character in the $$$i$$$-th line is $$$a_{i,j}$$$. If $$$a_{i,j} = $$$ "*", then the $$$j$$$-th square in the $$$i$$$-th row should be black, and if $$$a_{i,j} = $$$ ".", then the $$$j$$$-th square in the $$$i$$$-th row should be white.

It's guaranteed that the sum of $$$n \times m$$$ over all testcases doesn't exceed $$$10^5$$$.

Output format:
For each testcase print a single integer — the smallest total price of the tiles needed to cover all the white squares in burles.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first testcase you are required to use a single $$$1 \times 1$$$ tile, even though $$$1 \times 2$$$ tile is cheaper. So the total price is $$$10$$$ burles.

In the second testcase you can either use two $$$1 \times 1$$$ tiles and spend $$$20$$$ burles or use a single $$$1 \times 2$$$ tile and spend $$$1$$$ burle. The second option is cheaper, thus the answer is $$$1$$$.

The third testcase shows that you can't rotate $$$1 \times 2$$$ tiles. You still have to use two $$$1 \times 1$$$ tiles for the total price of $$$20$$$.

In the fourth testcase the cheapest way is to use $$$1 \times 1$$$ tiles everywhere. The total cost is $$$6 \cdot 3 = 18$$$.

Example input:
4
1 1 10 1
.
1 2 10 1
..
2 1 10 1
.
.
3 3 3 7
..*
*..
.*.

Example output:
10
1
20
18

Editorial:
Notice that rows can be solved completely separately of each other. Each tile takes either one or two squares but it's always in the same row.
So let's take a look at a single row. There are sequences of dot characters separated by some asterisks. Once again each of these sequences can be solved independently of the others.
Thus, we have these empty strips of empty squares $$$1 \times k$$$ which, when solved, can be summed up into the whole answer.
There are two cases, depending on if a $$$1 \times 2$$$ is cheaper than two $$$1 \times 1$$$ tiles.
If it is then we want to use of many $$$1 \times 2$$$ tiles as possible. So given $$$k$$$, we can place $$$\lfloor \frac k 2 \rfloor$$$ $$$1 \times 2$$$ tiles and cover the rest $$$k - 2 \cdot \lfloor \frac k 2 \rfloor = k~mod~2$$$ squares with $$$1 \times 1$$$ tiles.
If it isn't cheaper then we want to cover everything with $$$1 \times 1$$$ tiles and never use $$$1 \times 2$$$ ones. So all $$$k$$$ should be $$$1 \times 1$$$.
The easier way to implement this might be the following. Let's update the price of the $$$1 \times 2$$$ tile with the minimum of $$$y$$$ and $$$2 \cdot x$$$. This way the first algorithm will produce exactly the same result of the second one in the case when a $$$1 \times 2$$$ tile isn't cheaper than two $$$1 \times 1$$$ ones.
Overall complexity: $$$O(nm)$$$ per testcase.