CodeForces445A DZY Loves Chessboard

A. DZY Loves Chessboard

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 and m (1 ≤ n, m ≤ 100).

Each of the next n lines contains a string of m characters: the j-th character of the i-th string is either "." or "-". A "." means that the corresponding cell (in the i-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 of m characters. The j-th character of the i-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

B

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.

MD，今天写这道题目的时候，想的太简单了，被一类数据搞得体无完肤。。。

Input

50 50
.........................--..................-....
............................................-.....
..-.....................................-.........
...........................-......................
........-.....................-...................
..................................................
.......-......-...................................
...-..................-.......................-...
.....-......-...............................-.....
...-................-...-..............

Output

BWBWBWBWBWBWBWBWBWBWBWBWB--BWBWBWBWBWBWBWBWBW-BWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBWWBWBWBWBWBWBWBWBW-BWBWB
BW-WBWBWBWBWBWBWBWBWBWBWBWBBWBWBWBWBWBWB-BWBWWBWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBW-BWBWBWBWBWBWBWBWBBWBWB
BWBWBWBW-WBWBWBWBWBWBWBWBWBWWB-BWBWBWBWBWBWBWWBWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBWBBWBWBWBWBWBWBWBWBBWBWB
BWBWBWB-BWBWBW-WBWBWBWBWBWBWWBWBWBWBWBWBWBWBWWBWBW
WBW-WBWBWBWBWBWBWBWBWB-BWBWBBWBWBWBWBWBWBWBWBB-BWB
BWBWB-BWBWBW-WBWBWBWBWBWBWBWWBWBWBWBWBWBWBWB-WBWBW
WBW-WBWBWBWBWBWBWBWB-BWB-BWBBWBWBWBWBWBWBWB...

Answer

BWBWBWBWBWBWBWBWBWBWBWBWB--WBWBWBWBWBWBWBWBWB-BWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB-BWBWB
BW-WBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBW-WBWBWBWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBW-WBWBWBWBWBWBWBWBWBWBWB
BWBWBWBW-WBWBWBWBWBWBWBWBWBWBW-WBWBWBWBWBWBWBWBWBW
WBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB
BWBWBWB-BWBWBW-WBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBW
WBW-WBWBWBWBWBWBWBWBWB-BWBWBWBWBWBWBWBWBWBWBWB-BWB
BWBWB-BWBWBW-WBWBWBWBWBWBWBWBWBWBWBWBWBWBWBW-WBWBW
WBW-WBWBWBWBWBWBWBWB-BWB-BWBWBWBWBWBWBWBWBW...

一开始想的就是把每个格子按顺序填写好就没问题吧，这就是问题，当然不对了，在有拐角的地方，按顺序，就是错误！！
好吧，Json的方法，就是在已经打好的答案里填写字母，就OK，（这种方法很变态啊，积累吧）

 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int max_size = ;

 int main()
 {
     int row, col;
     int graph[max_size][max_size];
     int vis[max_size][max_size];

     for(int i = ; i < max_size; i++)
     {
         for(int j = ; j < max_size; j++)
         {
             if((i+j) %  == )
                 graph[i][j] = ;
             else
                 graph[i][j] = ;
         }
     }
     scanf("%d %d%*c", &row, &col);

     for(int i = ; i < row; i++)
     {
         for(int j = ; j < col; j++)
         {
             char a = getchar();
             if(a == '-')
                 graph[i][j] = -;
         }
         getchar();
     }

     for(int i = ; i < row; i++)
     {
         for(int j = ; j < col; j++)
         {
             if(graph[i][j] == )
                 printf("B");
             else if(graph[i][j] == )
                 printf("W");
             else
                 printf("-");
         }
         puts("");
     }
     return ;
 }

代码