dzy loves chessboard

最新推荐文章于 2020-03-26 14:59:36 发布

zienyang

最新推荐文章于 2020-03-26 14:59:36 发布

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分类专栏：简单题文章标签： acm菜鸟

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本文介绍了一个棋盘布局问题的解决方案，目标是在好的格子上放置黑白棋子，并确保相邻格子的棋子颜色不同。通过一种简单的算法实现了这一目标，并提供了代码示例。

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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

简单题一道卡半年，acm渣渣连预处理都不知道，想了好几种推法一直都没办法ac，看了其他人的解法后感到了这个世界深深的恶意，呵呵呵呵呵呵呵~~~

#include<stdio.h>
#include<string.h>
int main( )
{
    int m,n,i,j;
    char a[100][100];
    char b[100][100];
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        a[0][0]='B';
        for(i=0;i<n;i++)
        {
            if(i%2==0)a[i][0]='B';
            else a[i][0]='W';
            for(j=1;j<n;j++)
            {
                if(a[i][j-1]=='B')a[i][j]='W';
                else if(a[i][j-1]='W')a[i][j]='B';
            }
        }
        for(i=0;i<m;i++)
            scanf("%s",b[i]);
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
                if(b[i][j]=='-')printf("-");
                else printf("%c",a[i][j]);
            printf("\n");
        }
    }
    return 0;
}