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copied!<p>What you want to do is a <a href="http://en.wikipedia.org/wiki/Eulerian_path" rel="nofollow">Eulerian Path</a>.. I had solved the same problem on <a href="http://www.codechef.com/problems/WORDS1" rel="nofollow">Codechef</a>. This is my Code if you wanna use.. Plz tell me if you need a explanation,it is very easy to understand though.</p> <pre><code>#include <iostream> #include <string.h> #include <string> using namespace std; int visit[26]; int adj[26][26]; int count=0; void scc(int i) //Strongly COnnected Component { visit[i]=-1;//visiting for(int t=0;t<26;t++) { if(adj[i][t]>0 && visit[t]==0)//not visited yet scc(t); } visit[i]=1; count++; } int main() { string in; int t,n,k,nv,counta,countb,flag; int a[26],b[26]; cin >> t; while(t--) { cin >> n; memset(a,0,26*sizeof(int)); memset(b,0,26*sizeof(int)); memset(visit,0,26*sizeof(int)); memset(adj,0,26*26*sizeof(int)); k=26;count=0;counta=0;countb=0;flag=0;nv=0; while(n > 0) { n--; cin >> in; a[in[0]-'a']++; b[in[in.size()-1]-'a']++; adj[in[0]-'a'][in[in.size()-1]-'a'] = 1; adj[in[in.size()-1]-'a'][in[0]-'a'] = 1; } for(int i=0;i<26;i++) if(a[i]>0) { scc(i); break; } for(int i=0;i<26;i++) if(a[i]!=0 || b[i]!=0) nv++; if(count!=nv) flag=1; while(k > 0 && flag!=1 ) { if(a[k-1]-b[k-1] == 1) counta++; else if(b[k-1]-a[k-1] == 1) countb++; else if(a[k-1]!=b[k-1]) flag = 1; k--; } if(flag==0 && counta==countb && ( counta==1 || counta ==0)) cout << "Ordering is possible." <<endl; else cout << "The door cannot be opened." <<endl; } return 0; } </code></pre>

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