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11550 - Demanding Dilemma
Posted: Mon Nov 03, 2008 4:05 am
by rij
pls give me some test cases.i m getting wa.
heres my code
Re: 11550 - Demanding Dilemma
Posted: Mon Nov 03, 2008 2:27 pm
by 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...
Re: 11550 - Demanding Dilemma
Posted: Mon Nov 03, 2008 2:54 pm
by mmonish
try this case
Input:
My AC output:
hope this helps..
Re: 11550 - Demanding Dilemma
Posted: Mon Nov 03, 2008 7:39 pm
by rij
corrected but still
wa
Code: Select all
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;
}
Re: 11550 - Demanding Dilemma
Posted: Mon Nov 03, 2008 7:45 pm
by Vytenis
You must also check if there are no two identical edges defined by the incidence matrix. If there are, you should return "No".
Re: 11550 - Demanding Dilemma
Posted: Tue Nov 04, 2008 11:49 am
by rij
thanks got ac after a number of tries.
Re: 11550 - Demanding Dilemma
Posted: Wed Nov 05, 2008 6:34 am
by zhouerjin
rij wrote:thanks got ac after a number of tries.
can you say more detail?
I still WA...
Re: 11550 - Demanding Dilemma
Posted: Wed Nov 05, 2008 8:12 am
by SerailHydra
The graph should be a simple indirected one.
Pay attention to the word 'simple'.
Re: 11550 - Demanding Dilemma
Posted: Thu Nov 06, 2008 6:23 am
by zhouerjin
I still WA....I need help...
Code: Select all
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;
}
Re: 11550 - Demanding Dilemma
Posted: Thu Jan 14, 2010 6:21 pm
by shakil
To : zhouerjin
Change only one line & i hope that will make you AC.
change to ->
Because if u & v two are -1 then u*v>0
Re: 11550 - Demanding Dilemma
Posted: Fri Aug 17, 2012 4:31 am
by @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 :
Code: Select all
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 :
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)
Re: 11550 - Demanding Dilemma
Posted: Tue Dec 11, 2012 9:26 am
by brianfry713
Re: 11550 - Demanding Dilemma
Posted: Thu Apr 16, 2015 1:01 pm
by uDebug
Added some input to help with testing / debugging here:
http://www.udebug.com/UVa/11550