11550 - Demanding Dilemma

[rij]

pls give me some test cases.i m getting wa.
heres my code

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[zhouerjin]

I WA too...-_-!
but I can't understand your code on:if(arr[j]=='1'&& i!=j)count++;
why do you check if i!=j? I think it doesn't need...

[mmonish]

try this case
Input:

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1
3 2
1 1
1 0
0 1

My AC output:

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Yes

hope this helps..

[rij]

corrected but still wa

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void call(int n,int m){
		int count=0,fc=0;
		for(int j=0;j<m;j++){
			count=0;
			for(int i=0;i<n;i++){
				if(arr[i][j]=='1')count++;
			}
			if(count==2)
				fc++;
		}
		if(fc==m)
			cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
		return;
}

[Vytenis]

You must also check if there are no two identical edges defined by the incidence matrix. If there are, you should return "No".

[rij]

thanks got ac after a number of tries.

[zhouerjin]

thanks got ac after a number of tries.

can you say more detail?
I still WA...

[SerailHydra]

The graph should be a simple indirected one.

Pay attention to the word 'simple'.

[zhouerjin]

I still WA....I need help...

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void prepare()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
		flag[i][j]=false;
	for (int i=1;i<=n;i++)
	for (int j=1;j<=m;j++)
		scanf("%d",&map[i][j]);
}

bool work()
{
	for (int i=1;i<=m;i++)
	{
		int u=-1,v=-1;
		for (int j=1;j<=n;j++)
		if (map[j][i])
		{
			if (u==-1) u=j;else 
			if (v!=-1) return false;else v=j;
		}
		if (u*v<0) return false;
		if (flag[u][v]) return false;
		flag[u][v]=flag[v][u]=true;
	}
	return true;
}

[shakil]

To : zhouerjin
Change only one line & i hope that will make you AC.

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if (u*v<0) return false;

change to ->

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if (u==-1||v==-1) return false;

Because if u & v two are -1 then u*v>0

[@li_kuet]

This is a very simple problem.
Just check if every column(edge) has exactly two vertex and no two columns(edges) has same vertexes(Because multiple edges are not allowed)
You can try this Cases
Input :

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3

3 3
1 1 0
1 1 1
1 0 0

2 2
1 1
1 1

3 3
1 0 0
1 1 1
0 1 1

Output :

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No
No
No

In 1st case first column(edge) has three vertex which is not possible
In 2nd case 1st and 2nd edges have the same vertexes (1,2) and (1,2) so again No
In 3rd case same thing occurs like 2nd case.Here 2nd and 3rd edges have the same vertexes (2,3) and (2,3)

[brianfry713]

input

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1
4 3
1 1 1
1 0 0
0 1 0
0 0 0

output No

[uDebug]

Added some input to help with testing / debugging here:

http://www.udebug.com/UVa/11550