Fill the Square
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Description
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 filledfor 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:
1. Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common
edge.
2. 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
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 theinput matrix with the empty cells filled heeding the rules above.
Sample Input
23
...
...
...
3
...
A..
...
Case 1:
ABA
BAB
ABA
Case 2:
BAB
ABA
BAB
既然是字典序之和最小,那么我们只需一个一个位置填充最小的字母即可。在输入正方形网格时,我使用的之前学过的函数输入。代码如下:
#include<cstdio>#include<iostream>using namespace std;int n;char A[11][11];char zimu(){char s;for(;;){s=getchar();if(s=='.'||(s>='A'&&s<='Z')) return s; }}int main(){int T,count=0;cin>>T;while(T--){cin>>n;for(int i=0;i<n;i++)for(int j=0;j<n;j++)A[i][j]=zimu();for(int i=0;i<n;i++){for(int j=0;j<n;j++){if(A[i][j]!='.') continue;char z='A';while(z<=90){A[i][j]=z++;if(i>0&&(A[i][j]==A[i-1][j])) continue;if(i<(n-1)&&(A[i][j]==A[i+1][j])) continue;if(j>0&&(A[i][j]==A[i][j-1])) continue;if(j<(n-1)&&(A[i][j]==A[i][j+1])) continue;break;}}}printf("Case %d:\n",++count);for(int i=0;i<n;i++){for(int j=0;j<n;j++)printf("%c",A[i][j]);printf("\n");}}return 0;}
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