DZY Loves Chessboard

最新推荐文章于 2020-03-26 14:59:36 发布

转载最新推荐文章于 2020-03-26 14:59:36 发布 · 89 阅读

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原文链接：http://www.cnblogs.com/sylvialucy/p/4119130.html

本文介绍了一个棋盘布局问题的解决方案，目标是在给定的好坏格子中放置黑白棋子，确保相邻格子的棋子颜色不同。通过预先填充棋盘并根据好坏格子进行调整，实现了有效的布局方案。

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Description
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 Input
Input
1 1
.
Output
B
Input
2 2
..
..
Output
BW
WB
Input
3 3
.-.
---
--.
Output
B-B
---
--B
Hint
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.

 1 //这是一道bfs题，但是也可以当成一道模拟题。。。我目前算法比较渣，但是想到了这种方法
 2 //先把格子填满，输入'.'时只要把对应填好的B 或 W替换掉就可以了
 3 #include <stdio.h>
 4 #include <string.h>
 5 char map[102][102];
 6 char in[102][102];
 7 
 8 int main()
 9 {
10     int i,j,m,n,t;
11     for(i=0;i<102;i++)
12     {
13         for(j=0;j<102;j++)
14         {
15             if((i+j)%2)map[i][j]='B';
16             else map[i][j]='W';
17         }
18     }
19     while(~scanf("%d %d",&n,&m))
20     {
21         getchar();
22         memset(in,0,sizeof (in));
23         for(i=0;i<n;i++)
24         {
25             gets(in[i]);
26         }
27         for(i=0;i<n;i++)
28         {
29             for(j=0;j<m;j++)
30             {
31                 if(in[i][j]=='.')in[i][j]=map[i][j];
32             }
33         }
34         for(i=0;i<n;i++)
35             puts(in[i]);
36     }
37     return 0;
38 }

转载于:https://www.cnblogs.com/sylvialucy/p/4119130.html