DZY Loves Chessboard （周赛2）

最新推荐文章于 2020-07-09 18:48:32 发布

原创最新推荐文章于 2020-07-09 18:48:32 发布 · 456 阅读

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本文介绍了一个简单的算法，用于解决在给定的棋盘上放置黑白棋子的问题，确保相邻格子的棋子颜色不同。通过将棋盘上的好位置按照行列坐标之和的奇偶性分别放置黑色和白色棋子，可以有效避免相同颜色的棋子相邻。

DZY loves chessboard, and he enjoys playing with it.

He has a chessboard of n rows and m columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.

You task is to find any suitable placement of chessmen on the given chessboard.

Input

The first line contains two space-separated integers n andm (1 ≤ n, m ≤ 100).

Each of the next n lines contains a string ofm characters: the j-th character of thei-th string is either "." or "-". A "." means that the corresponding cell (in thei-th row and the j-th column) is good, while a "-" means it is bad.

Output

Output must contain n lines, each line must contain a string ofm characters. The j-th character of thei-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.

If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.

Sample test(s)

Input

1 1
.

Output

Input

2 2
..
..

Output

BW
WB

Input

3 3
.-.
---
--.

Output

B-B
---
--B

Note

In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.

In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.

In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.

水题不解释上代码。

#include <stdio.h>
#include <string.h>

int main()
{
    int n,m;
    char a[100][100];
    while(~scanf("%d%d",&n,&m))
    {
        int i,j;
        for(i=0;i<n;i++)
        {
            scanf("%s",a[i]);
        }


            for(i=0;i<n;i++)
            {
                 for(j=0;j<m;j++)
                {

                        if(a[i][j]=='.' && (i+j)%2==0)
                        a[i][j]='B';
                        else if(a[i][j]=='.' && (i+j)%2!=0)
                        a[i][j]='W';
                }

            }
                for(i=0;i<n;i++)
                {
                    for(j=0;j<m;j++)
                        printf("%c",a[i][j]);

                        printf("\n");
                }

    }
    return 0;
}