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B. New Theatre Square

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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×m meters. However, the picture is about to get more complicated now. Let ai,j be the j-th square in the i-th row of the pavement.

You are given the picture of the squares:

if ai,j= “*”, then the j-th square in the i-th row should be black;

if ai,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:

1×1 tiles — each tile costs x burles and covers exactly 1 square;

1×2 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 1×1 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

The first line contains a single integer t (1≤t≤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≤n≤100; 1≤m≤1000; 1≤x,y≤1000) — the size of the Theatre square, the price of the 1×1 tile and the price of the 1×2 tile.

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

It’s guaranteed that the sum of n×m over all testcases doesn’t exceed 105.

Output

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

Example

inputCopy 4 1 1 10 1 . 1 2 10 1 … 2 1 10 1 . . 3 3 3 7 …* …

.. outputCopy 10 1 20 18 Note In the first testcase you are required to use a single 1×1 tile, even though 1×2 tile is cheaper. So the total price is 10 burles.

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

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

In the fourth testcase the cheapest way is to use 1×1 tiles everywhere. The total cost is 6⋅3=18.

题意： 给你n * m的方格， * 代表黑色格子 . 代表白色格子，其中黑色格子已经确定，现有一些 1 * 1 和 1 * 2的白色格子，且不能旋转，花费分别为x和y，问最少花费多少钱

思路：

• 若x<y，全都铺设1 * 1的花费最少
• 若x>=y，则尽量铺设 1 * 2的，其次是1 * 1的

然后一行一行的遍历就好了

代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         using namespace std;char a[101][1001];int main(){ int t; cin >> t; while (t--) { int n, m, x, y; cin >> n >> m >> x >> y; int s = 0, d = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { cin >> a[i][j]; } for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { if (a[i][j] == '.') { if (j == m) { d++; } else { if (a[i][j + 1] == '.') { s++; j++; } else d++; } } } int ans = d * x; if (y >= x * 2) { ans += s * 2 * x; } else ans += s * y; cout << ans << endl; } return 0;}
        
       
      
     
    
   

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