Fill the Square

作者: Rememberautumn 分类: uva 发布时间: 2014-07-27 16:36 阅读: 1,770

题目大意:在一个n*网格中填了一些大写字母,你的任务是把剩下的格子也填满大写字母,使得任意两个相邻格子的字母不同,如有多种填法,填字典序最小的一组。如有疑问,请留言。

Fill the Square
Input: Standard Input

Output: Standard Output

 

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:

 

  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
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

Problemsetter: Sohel Hafiz

[codesyntax lang=”c++”]

#include<iostream>
using namespace std;
int main()
{
char a[15][15],ch;
bool l;
int n,i,t,j,k;
cin>>t;
for(k=1;k<=t;k++)
{cin>>n;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(a[i][j]==’.’)
{
for(ch=’A’;ch<=’Z’;ch++)
{
l=1;
if(i>0&&a[i-1][j]==ch)
l=0;
if(i<n-1&&a[i+1][j]==ch)
l=0;
if(j>0&&a[i][j-1]==ch)
l=0;
if(i<n-1&&a[i][j+1]==ch)
l=0;
if(l)
{a[i][j]=ch;break;}

 

}

}
cout<<“Case “<<k<<‘:'<<endl;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
cout<<a[i][j];

cout<<endl;
}

}

return 0;
}

[/codesyntax]

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