@tendua.....ya i was wrong i got it...
2011/7/15 ●αηυяαg ∩ ℓιƒє ≈ Φ <[email protected]>
> #include<iostream>
> #include<vector>
> #include<cstring>
>
> using namespace std;
>
> bool mark[20];
> vector<int> v[20];
> int count=0,n;
>
> void solve(int row)
> {
> if(row==n)count++;
>
> else
> for(int i=0;i<v[row].size();i++)
> {
> if(!mark[v[row][i]])
> {
> mark[v[row][i]]=true;
> solve(row+1);
> mark[v[row][i]]=false;
> }
> }
> }
>
> int main()
> {
> int x;
> cin>>n;
> for(int i=0;i<n;i++)
> for(int j=0;j<n;j++)
> {
> cin>>x;
> if(x)v[i].push_back(j);
> }
>
> memset(mark,false,sizeof(mark));
> solve(0);
>
> cout<<count<<endl;
> }
>
>
> On Jul 14, 11:23 pm, shilpa gupta <[email protected]> wrote:
> > if we have to find out no of possible outputs ...........then we can do
> this
> > for(i=1;i<=n;i++)
> > {
> > ans=ans * no of 1s in ith row
> > return(ans);}
> >
> > On Thu, Jul 14, 2011 at 11:46 PM, SkRiPt KiDdIe <[email protected]
> >wrote:
> >
> >
> >
> > > Will a Back-tracking solution suffice..??
> >
> > > --
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> >
> > --
> > shilpa gupta
> > b tech 2nd year
> > computer science and engineering
> > mnnit allahabad
>
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--
shilpa gupta
b tech 2nd year
computer science and engineering
mnnit allahabad
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