In this problem, you have to draw a square using uppercase English Alphabets.
To be more precise, you will be given a square grid with some empty blocks and others already filled for you with some letters to make your task easier. You have to insert characters in every empty cell so that the whole grid is filled with alphabets. In doing so you have to meet the following rules:
Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common edge. There could be many ways to fill the grid. You have to ensure you make the lexicographically smallest one. Here, two grids are checked in row major order when comparing lexicographically.
Input The first line of input will contain an integer that will determine the number of test cases. Each case starts with an integer n( n<=10 ), that represents the dimension of the grid. The next n lines will contain n characters each. Every cell of the grid is either a ‘.’ or a letter from [A, Z]. Here a ‘.’ Represents an empty cell.
Output For each case, first output Case #: ( # replaced by case number ) and in the next n lines output the input matrix with the empty cells filled heeding the rules above. Sample Input Output for Sample Input
2
3
...
...
...
3
...
A..
...
Case 1:
ABA
BAB
ABA
Case 2:
BAB
ABA
BAB
题意：给出一个填有大写字母的正方形矩阵，要求在剩余空格内填上大写字母使得每个字母相邻位置不同，有多种解时，字典序要求最小的一个。
思路：直接从A开始遍历即可，字典序最小从A开始遍历到Z时肯定为最小。
#include <iostream> using namespace std; #define MAXN 15 char grid[MAXN][MAXN]; int main() { int T, n; cin>>T; int cas = 1; for (int k = 0; k < T; k++) { cin>>n; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) cin>>grid[i][j]; for (i = 0; i < n; i++) for (int j = 0; j < n; j++) if (grid[i][j] == '.') { for (char ch = 'A'; ch <= 'Z'; ch++)//小到大 { int flag = true; if (i > 0 && grid[i - 1][j] == ch) flag = false; if (i < n - 1 && grid[i + 1][j] == ch) flag = false; if (j > 0 && grid[i][j - 1] == ch) flag = false; if (j < n - 1 && grid[i][j + 1] == ch) flag = false; if (flag) { grid[i][j] = ch; break; } } } printf("Case %d:\n", cas ++); for (i = 0; i < n; i++) { for (int j = 0; j < n; j++) cout<<grid[i][j]; cout<<endl; } } return 0; }
转载于:https://www.cnblogs.com/riasky/p/3464778.html
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