11110 - Equidivisions

[Vexorian]

This one gave me a lot of trouble in the contest, it seemed really easy but I couldn't stop getting a WA, and even reread the prompt finding out things I skipped, can anyone give me some tricky input/output samples I could try on my algorithm?

[david]

I also had a lot of trouble with this problem. The trick is that some cells appear twice in the input.

[vinay]

ok my tle is gone by concidering that fact....

but now WA...

i used union by rank method to make sets of each number....

if the number of elemnts in each set is not n then wrong...

but i m getting WA...

here is the code...

Code: Select all

ACC   :D 

pleae help me. [/code]

[Vexorian]

I also had a lot of trouble with this problem. The trick is that some cells appear twice in the input.

That's broken, cause the statement says that the input is a partition of the matrix, then that shouldn't happen, thanks though.

Edit: Now that I think about it, my alg should be able to resist that kind of input. I also tested with repeated cells and it was giving the correct answer

[aniket]

i cant find why wa
my algo is just simple
just cover all by default
than collect all the partition with their relative number
and then check each partition from the begining cell if it contains the given number of cells adajacent to it or not
here goes my code

Code: Select all

#include<stdio.h>

long mat[105][105],total_number,total[105],no_x[105],no_y[105],num;


int check(long x,long y,long sym)
{
	if(mat[x][y]!=sym)
		return 0;
	total_number++;
	mat[x][y]=-1;

	if(x-1>=1)
		check(x-1,y,sym);
	if(x+1<=num)
		check(x+1,y,sym);
	
	if(y-1>=1)
		check(x,y-1,sym);
	if(y+1<=num)
		check(x,y+1,sym);
return 0;
}

void main()
{
	long i,j,k,x,y,p,q,temp_count;
	char c;

	while(scanf("%ld",&num),num!=0)
	{
		for(i=1;i<=num;i++)
			for(j=1;j<=num;j++)
				mat[i][j]=1;

		for(i=2,p=0;i<=num;i++)
		{
			scanf("%ld%c",&x,&c);
			k=0;
			temp_count=1;
			while(c==' ')
			{
				scanf("%ld%c",&y,&c);
				temp_count++;
				if(temp_count==2)
				{
					if(mat[x][y]!=1)
						p=1;
					mat[x][y]=i;
					no_x[i]=x;
					no_y[i]=y;
					k++;
					temp_count=0;
				}
				else if(temp_count==1)
				{
					x=y;
				}

			}
			total[i]=k;
		}
		if(p==1)
		{printf("wrong\n");continue;}


		for(i=2,p=1;i<=num;i++)
		{
			total_number=0;
			check(no_x[i],no_y[i],i);
			if(total_number!=total[i])
			{p=0;printf("wrong\n");break;}
		}

		if(p==1)
		{
			//{p=0;printf("wrong\n");break;}
			for(i=1,q=0;i<=num;i++)
			{	for(j=1;j<=num;j++)
				{
					if(mat[i][j]==1)
					{q=1;break;}
				}
				if(q==1)
					break;
			}

			if(q==0)
				printf("good\n");

			else
			{
				no_x[1]=i;no_y[1]=j;
				total[1]=0;
				for(i=1,q=0;i<=num;i++)
					for(j=1;j<=num;j++)
						if(mat[i][j]==1)
							total[1]++;

				total_number=0;
				check(no_x[1],no_y[1],1);

				if(total_number!=total[1])
					printf("wrong\n");
				else
					printf("good\n");
			}


		}
	}

}

Hope any kind helper wish to compile and check the code for rescuing me
bye

[Darko]

I didn't check that code above, but I got frustrated with this problem, too.

My mistake was that I had a line like:

Code: Select all

// if not exactly n pairs of numbers on a line, it's wrong
if (st.countTokens() != 2 * n) ok = false;

I changed that to "less than" and it worked - same cell might appear in the description of the *single* partition more than once. I am guessing that is what david meant (I understood it as "same cell can appear in different partitions" - which also might be true).

[sohel]

I was really frustrated with this problem in the online contest..

I also assumed if ( v.size() != 2*n ) ok = false .

And another thing: there were more than one value with the same coordinate... now, which do u consider??? General coding would tend to consider that last one as it would overwrite the earlier ones and it did work... but there is no clear explanation as to why we have to consder the last one.

Poorly written problemstatement....

[Vexorian]

I see! A partition is not necessarily going to have n points in each partition input.

[Sedefcho]

So what is the fix for this "no problem"-problem

[mrahman]

Hi,
aniket, Your approach is good. But your Input taking method failed you to accept this code. Try to use gets() function instead of scanf() function and then parse the integer from the string. and you could simplify your code like

Code: Select all

#include<stdio.h> 

long mat[105][105],total_number,total[105],no_x[105],no_y[105],num; 


int check(long x,long y,long sym) 
{ 
   if(mat[x][y]!=sym) 
      return 0; 
   total_number++; 
   mat[x][y]=-1; 

   if(x-1>=1) 
      check(x-1,y,sym); 
   if(x+1<=num) 
      check(x+1,y,sym); 
    
   if(y-1>=1) 
      check(x,y-1,sym); 
   if(y+1<=num) 
      check(x,y+1,sym); 
return 0; 
} 

void main() 
{ 
   long i,j,k,x,y,p,q,temp_count; 
   char c; 

   while(scanf("%ld",&num),num!=0) 
   { 
      for(i=1;i<=num;i++) 
         for(j=1;j<=num;j++) 
            mat[i][j]=1; 
	  total[1] = num*num;	//add this line
	  for(i=1;i<=num;i++)		  
	  {
		  no_x[i] = -1;
		  no_y[i] = -1;
	  }
      for(i=2,p=0;i<=num;i++) 
      { 
         scanf("%ld%c",&x,&c); 
         k=0; 
         temp_count=1; 
         while(c==' ') 
         { 
            scanf("%ld%c",&y,&c); 
            temp_count++; 
            if(temp_count==2) 
            { 
				//remove this
               /*if(mat[x][y]!=1) 
                  p=1; */
               mat[x][y]=i; 
               no_x[i]=x; 
               no_y[i]=y; 
               k++; 
               temp_count=0; 
            } 
            else if(temp_count==1) 
            { 
               x=y; 
            } 

         } 
         total[i]=k; 
		 total[1]-=k; //add this line
      } 
	  //remove this
      /*if(p==1) 
      {printf("wrong\n");continue;} */
	  // Add below's code for initialize
	  p=0;
		for(i=1;i<=num && !p;i++) 
         for(j=1;j<=num && !p;j++) 
		 {
			 if( mat[i][j]==1){no_x[1] = i; no_y[1] = j; p = 1;}
		 }
		 // End

      for(i=1,p=1;i<=num;i++)  
      { 
         if(no_x[i]!=-1 && no_y[i]!=-1)  // Add this checking
		 {
			total_number=0; 
         
			check(no_x[i],no_y[i],i); 
			if(total_number!=total[i]) 
				{p=0;printf("wrong\n");break;} 
		 }
      } 
	   if(p)printf("good\n"); //add this output line	
	  //remove this
	/*
      if(p==1) 
      { 
         //{p=0;printf("wrong\n");break;} 
         for(i=1,q=0;i<=num;i++) 
         {   for(j=1;j<=num;j++) 
            { 
               if(mat[i][j]==1) 
               {q=1;break;} 
            } 
            if(q==1) 
               break; 
         } 

         if(q==0) 
            printf("good\n"); 

         else 
         { 
            no_x[1]=i;no_y[1]=j; 
            total[1]=0; 
            for(i=1,q=0;i<=num;i++) 
               for(j=1;j<=num;j++) 
                  if(mat[i][j]==1) 
                     total[1]++; 

            total_number=0; 
            check(no_x[1],no_y[1],1); 

            if(total_number!=total[1]) 
               printf("wrong\n"); 
            else 
               printf("good\n"); 
         } 


      } */
   } 

} 

Good Luck!

Sorry for my poor english.

[Vexorian]

My mistake besides of expecting each line to have n pairs was that it was possible for a number to exceed n or be 0 , the problem statement says it clearly, the numbers will be non-negative numbers, never said anything about the numbers being in the correct range.

[slxst]

Could someone provide tricky test cases?

I checked all the possible flaws here discussed but my code doesn't fall on those. However I still get WA!

Thanks!

[mrahman]

Hi slxst,

Try this

input #

Code: Select all

2
1 2 0 0
5
0 0 1 2 1 3 3 2 2 2
2 1 4 2 4 1 5 1 3 1 3
4 5 5 2 5 3 5 5 5 4
2 5 3 4 3 5 4 3 4 4
0

output #

Code: Select all

good
wrong

Hope it will help.

[helloneo]

Hi slxst,
Try this

input #

Code: Select all

5
0 0 1 2 1 3 3 2 2 2
2 1 4 2 4 1 5 1 3 1 3
4 5 5 2 5 3 5 5 5 4
2 5 3 4 3 5 4 3 4 4

I strongly doubt that that kind of input come..
My AC program can't handle that..

[rio]

still WA...

Hi slxst,

Try this

input #

Code: Select all

2
1 2 0 0
5
0 0 1 2 1 3 3 2 2 2
2 1 4 2 4 1 5 1 3 1 3
4 5 5 2 5 3 5 5 5 4
2 5 3 4 3 5 4 3 4 4
0

output #

Code: Select all

good
wrong

Hope it will help.

i can't understand why the first case is "good".
if theres an invalid cell donate, like 0 0, i'm considering it "wrong"

-----
sorry for my poor english. ('A`)