11585 - Nurikabe

[Igor9669]

Please explain me why in second example the answer is "not solved"???

3#1#2
.###.
.#1##
####3
2.#..

[SerailHydra]

"For any unshaded space b, there is a sequence of unshaded spaces starting at b and ending at a space on the edge of the grid where consecutive spaces in this sequence share a side or a corner. "

[Igor9669]

I understood, that I don't understand this statement !
Could you please show me in the figure above, where is a mistake...

[sohel]

Code: Select all

3#1#2
.###.
.#1##
####3
2.#..

Look at the cell with coordinates (2,2) - there is no way to reach the edges of the grid following only unshaded spaces.

[vahid sanei]

Sohel you got Acc in 0.170 sec
i use bfs for this problem but i got acc in 0,540 , how can i reduce time of my code ?



thanks in advance

[sohel]

I used dfs().
May be bfs() codes has some overheads. Another reason could be the extensive use of STLs. STL codes are inherently slow!
The time difference isn't actually that significant to cause any worry . A high constant factor could be the reason for the difference in time.
You can also speed up by using efficient methods for reading the input.

[vahid sanei]

what is faster than scanf and printf ?

[sohel]

fread()

[Igor9669]

I used dfs, but still WA....

Please anybody give some tests!
I don't know where I have a mistake...

What will be the answer for that test:

3

1 2 0
##

1 1 0
.

1 1 0
#

[vahid sanei]

Input

Code: Select all

4
1 2 0
##
1 1 0
.
1 1 0
#
1 2 1
0 0 1
.#

Output

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not solved
not solved
not solved
solved

[Igor9669]

Finally accepted!!!!!
But my ac prog gives such answer on my test:

Code: Select all

3

1 2 0
##

1 1 0
.

1 1 0
#

Code: Select all

solved
not solved
solved

[vahid sanei]

in every 2 x 2 subsquare there is at least one unshaded space.

those arent 2 x 2 subsquare but i checked

[Chirag Chheda]

I am getting WA for this problem. I cant find my mistake. Can someone help me??

Code: Select all

#include<iostream>

using namespace std;

int r,c,ans;
char arr[102][102],s[102];
bool cnt[102][102],fl;

void dfs1(int i,int j)
{
     cnt[i][j]=1;
     ans++;
     
     if(i-1>=0 && arr[i-1][j]=='#' && cnt[i-1][j]==0) dfs1(i-1,j);
     if(i+1<r && arr[i+1][j]=='#' && cnt[i+1][j]==0) dfs1(i+1,j);
     if(j-1>=0 && arr[i][j-1]=='#' && cnt[i][j-1]==0) dfs1(i,j-1);
     if(j+1<c && arr[i][j+1]=='#' && cnt[i][j+1]==0) dfs1(i,j+1);          
}

void dfs(int i,int j)
{
     cnt[i][j]=1;
     ans++;

     if(i-1>=0 && arr[i-1][j]=='.' && cnt[i-1][j]==0) dfs(i-1,j);
     if(i+1<r && arr[i+1][j]=='.' && cnt[i+1][j]==0) dfs(i+1,j);
     if(j-1>=0 && arr[i][j-1]=='.' && cnt[i][j-1]==0) dfs(i,j-1);
     if(j+1<c && arr[i][j+1]=='.' && cnt[i][j+1]==0) dfs(i,j+1); 
}

void dfs2(int i,int j)
{
     if(i==0 || j==0 || i+1==r || j+1==c)
     fl=1;
     
     cnt[i][j]=1;
     
     if(i-1>=0 && arr[i-1][j]=='.' && cnt[i-1][j]==0) dfs2(i-1,j);
     if(i+1<r && arr[i+1][j]=='.' && cnt[i+1][j]==0) dfs2(i+1,j);
     if(j-1>=0 && arr[i][j-1]=='.' && cnt[i][j-1]==0) dfs2(i,j-1);
     if(j+1<c && arr[i][j+1]=='.' && cnt[i][j+1]==0) dfs2(i,j+1);
     
     if(i-1>=0 && j-1>=0 && arr[i-1][j-1]=='.' && cnt[i-1][j-1]==0) dfs2(i-1,j-1);
     if(i+1<r && j-1>=0 && arr[i+1][j-1]=='.' && cnt[i+1][j-1]==0) dfs2(i+1,j-1);
     if(i+1<r && j+1<c && arr[i+1][j+1]=='.' && cnt[i+1][j+1]==0) dfs2(i+1,j+1);
     if(i-1>=0 && j+1<c && arr[i-1][j+1]=='.' && cnt[i-1][j+1]==0) dfs2(i-1,j+1);      
}

int main()
{
    int t,i,j,k,e[50002][3],ans1,d;
    bool f;
    scanf("%d",&t);
    
    while(t--)
    {
        scanf("%d %d %d",&r,&c,&d);
        memset(cnt,0,sizeof(cnt));
        
        for(i=0;i<d;i++)
        scanf("%d %d %d",&e[i][0],&e[i][1],&e[i][2]);
        
        gets(s);
        for(i=0;i<r;i++)
        gets(arr[i]);
        
        ans=0;
        ans1=0;
        f=0;
        for(i=0;i<r;i++) 
        for(j=0;j<c;j++)
        {
            if(arr[i][j]=='#' && f==0)
            {
                dfs1(i,j);
                f=1;
            }
            if(arr[i][j]=='#')
            ans1++;
        }
        
        if(ans1!=ans)
        {
            puts("not solved");
            continue;
        }
        
        memset(cnt,0,sizeof(cnt));
        
        for(i=0;i<d;i++)
        {
            ans=0;
            dfs(e[i][0],e[i][1]);
            
            if(ans!=e[i][2])
            break;
        }
        
        if(i!=d)
        {
            puts("not solved");
            continue;
        }
        
        f=0;
        for(i=0;i<r-1;i++)
        {
            for(j=0;j<c-1;j++)
            {
                 if(arr[i][j]=='#' && arr[i+1][j]=='#' && arr[i+1][1+j]=='#' && arr[i][j+1]=='#')
                 {
                   f=1;
                   break;
                 }
            }
            
            if(f) break;
        }
        
        if(f)
        {
            puts("not solved");
            continue;
        }
        
        memset(cnt,0,sizeof(cnt));
        
        f=0;
        for(i=0;i<r;i++)
        {
              for(j=0;j<c;j++)
              {
                    if(arr[i][j]=='.' && cnt[i][j]==0)
                    {
                       fl=0;
                       dfs2(i,j);
                       if(fl==0)
                       {
                       f=1;
                       break;
                       }
                    }
              }
              if(f)
              break;
        }
        
        if(f)
        {
            puts("not solved");
            continue;
        }
        
        else
        {
            puts("solved");
        }
    }
    system("pause");
    return 0;
}

[Chirag Chheda]

I have a doubt. Can thr be such type of input :-

Code: Select all

1
3 4 2
0 3 1
1 1 1
###.
#.##
.##.

the description tells that there are only 2 dots but there are 4 in the actual nurikabe.
If yes what should be the answer

[vahid sanei]

Code: Select all

not solved