uva 11520 Fill the Square（DFS）

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:
 
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

题目大意：给出一个表，按表的要求生成一个图，并且相邻的格子字母不能相同，而且要求字典最小。
解题思路：DFS遍历并且判断。

#include<iostream>

using namespace std;

int  drg[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
char map[10][10];
int n;

void buile_map();
void drw_map();
int find(int x,int y,char c);

int main()
{
int t;	cin>>t;

for(int k=1;k<=t;k++)
{
cin>>n;

buile_map();

cout<<"Case "<<k<<":"<<endl;

drw_map();
}
return 0;}

void buile_map()
{
//	memset(map,'0',sizeof(map));

for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>map[i][j];
}

void drw_map()
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(map[i][j]=='.')
{	char c='A';
for(;find(i,j,c);c++);
map[i][j]=c;
}

cout<<map[i][j];
}
cout<<endl;
}
}

int find(int x,int y,char c)
{
int p,q;
for(int k=0;k<4;k++)
{
p=x+drg[k][0];
q=y+drg[k][1];

if(p<0||p>=n)	continue;
if(q<0||q>=n)	continue;
if(map[p][q]==c)return 1;
}
return 0;
}