求字典序最小解+贪心

Problem A
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.

 

OutputFor 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

思路：填的时候从A到Z一个一个试，是字典序最小的，不用担心这样下去最后不符合怎么办，因为每一个格子只需要考虑上下左右四个格子，有26个字母，肯定可以满足。

下面是代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXN=15;char grid[MAXN][MAXN];int N;int main(){    #ifndef ONLINE_JUDGE        freopen("in.txt","r",stdin);    #endif    int t;    scanf("%d",&t);    for(int cas=1;cas<=t;cas++)    {        scanf("%d",&N);        for(int i=0;i<N;i++) scanf("%s",&grid[i]);        for(int i=0;i<N;i++)        {            for(int j=0;j<N;j++)            {                if(grid[i][j]=='.')                {                    for(char k='A';k<='Z';k++)                    {                        bool ok=true;                        if(i>0&&grid[i-1][j]==k) ok=false;                        if(i<N-1&&grid[i+1][j]==k)ok=false;                        if(j>0&&grid[i][j-1]==k) ok=false;                        if(j<N-1&&grid[i][j+1]==k)ok=false;                        if(ok)                        {                            grid[i][j]=k;                            break;                        }                    }                }            }        }        printf("Case %d:\n",cas);        for(int i=0;i<N;i++)        {            cout<<grid[i]<<endl;        }    }    return 0;}