zoukankan      html  css  js  c++  java
  • A. DZY Loves Chessboard

    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.

    Examples
    input
    Copy
    1 1
    .
    output
    Copy
    B
    input
    Copy
    2 2
    ..
    ..
    output
    Copy
    BW
    WB
    input
    Copy
    3 3
    .-.
    ---
    --.
    output
    Copy
    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.

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <unordered_set>
    #include <unordered_map>
    #define ll long long
    using namespace std;
    int dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
    int n, m;
    vector<string> s(n);
    void dfs(int x, int y, bool t)
    {
        s[x][y] = (t ? 'B' : 'W');
        for (int i = 0; i < 4; i++)
        {
            int nx = x + dir[i][0];
            int ny = y + dir[i][1];
            if (nx>=0 && ny>=0 && nx+1<=n && ny+1<=m && s[nx][ny]=='.')
                dfs(nx, ny, t ^ 1);
        }
    }
    int main()
    {
        cin >> n >> m;
        s.resize(n);
        for (int i = 0; i < n; i++)
            cin >> s[i];
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (s[i][j] == '.') dfs(i, j, 1);
            }
        }
        for (int i = 0; i < n; i++)
        {
            if (i == n - 1)
                cout << s[i];
            else
                cout << s[i] << endl;
        }
        //system("pause");
        return 0;
    }
  • 相关阅读:
    jQuery实现 自动滚屏操作
    jQuery实现全选、全不选以及反选操作
    读曾国藩
    把时间当作朋友 之感知时间
    把时间当作朋友4未知永远存在
    Android N 设置中语言列表介绍
    如何编译ICU资源
    idea常用快捷键
    shell 笔记
    Json笔记
  • 原文地址:https://www.cnblogs.com/dealer/p/12308347.html
Copyright © 2011-2022 走看看