SPOJ: WORDS1 - Play on Words (Eular Path)


SPOJ: WORDS1 - Play on Words (Eular Path)


Idea: 
For a given list of words simply checks whether we can form a chain of words joining 
end of word with begining of word. As we know that for a graph to have an eularian path, there 
must be either no odd degree vertices or exactly two vertices having odd degree,in this case 
one verices must have one extra outgoing degree and another must have one extra incoming 
degree. (Note: as a whole graph should be connected, check this by  union find.) 

///============================================================================///
/// ///
/// IT'S ME ///
/// BISHAL GAUTAM ///
/// CSE,JAHANGIRNAGAR UNIVERSITY,DHAKA ///
/// "Follow Excellence..Success will chase U" ///
/// ///
///============================================================================///
#include<bits/stdc++.h>
#define PI acos(-1.0)
#define X first
#define Y second
#define mpp make_pair
#define nl printf("\n")
#define SZ(x) (int)(x.size())
#define pb(x) push_back(x)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>

///==Scanning====///
#define S(a) scanf("%d",&a)
#define P(a) printf("%d",a)
#define SL(a) scanf("%lld",&a)
#define S2(a,b) scanf("%d%d",&a,&b)
#define S3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define SL2(a,b) scanf("%lld%lld",&a,&b)
#define SL3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)

///==Arr,Vec Functions==///
#define all(v) v.begin(),v.end()
#define CLR(a) memset(a,0,sizeof(a))
#define SET(a) memset(a,-1,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define MAX(a) (*max_element(all(a)))
#define MIN(a) (*min_element(all(a)))
#define fv(i,v) for(int i=0;i<(int)v.size();i++)
#define fr(i,a,n) for(int i=a;i<=n;i++)
#define rf(i,n,a) for(int i=n;i>=a;i--)

///===Some Useful====///
#define OnBit(a) __builtin_popcountll((ll)a)
#define LB(v,k) lower_bound(v.begin(),v.end(),k)
#define _cin ios_base::sync_with_stdio(0),cin.tie(0)
#define dbg(x) cerr<<__LINE__<< ":: "<<#x<<"= "<<x<<endl
#define UNIK(v) sort(all(v)),v.resize( unique(all(v)) -v.begin() );
#define fi(n) for(__typeof(n.begin()) it=n.begin();it!=n.end();it++)
#define IO freopen("A.in","r",stdin),freopen("Out.out","w",stdout)
using namespace std;

///===TypeDefs=====///
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<pii> vii;

///===Number Theory===///
//template< class T > inline T SQR(T a) { return ((a)*(a)); }
//template< class T > inline T gcd(T a,T b) {a=abs(a);b=abs(b); if(!b) return a; return gcd(b,a%b);}
//template< class T > inline T lcm(T a,T b) {a=abs(a);b=abs(b); return (a/_gcd(a,b))*b;}
//template<typename T> T POW(T b,T p) {T r=1; while(p){if(p&1)r=(r*b);b=(b*b);p>>=1;}return r;}
//template<typename T> T BigMod(T b,T p,T m) {T r=1; while(p){if(p&1)r=(r*b)%m;b=(b*b)%m;p>>=1;}return r;}
//template<typename T> T ModInverse(T n,T m) { return BigMod(n,m-2,m); }
//
/////==GeoMetry=========///
//double DEG(double x) { return (180.0*x)/(PI);}
//double RAD(double x) { return (x*(double)PI)/(180.0);}
//template<typename T> double DIS(T a,T b){ return sqrt((double)( SQR(a.X-b.X)+ SQR(a.Y-b.Y))); }
//template<typename T> T ANGLE(T a,T b){ return atan( double(a.Y-b.Y) / double(a.X-b.X));}
//template<typename T> int isLeft(T a,T b,T c) { return (c.X-a.X)*(b.Y-a.Y)-(b.X-a.X)*(c.Y-a.Y); }
//
/////===IO-Related===///
//template< class T > void prnt(T v) { fv(i,v) {if(!i)cout<<v[i];else cout<<" "<<v[i];} cout<<endl; }
//template< class T > void BIN(T n) { if(!n){nl;return;}BIN(n/2);cout<<(n%2); }

///====Some-Stuff===///
/// atoi( str.c_str() ); // char string to int
/// sprintf(str,"%d",num);// num to char string
///int month[]={-1,31,28,31,30,31,30,31,31,30,31,30,31}; //Not Leap Year
///int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
///int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 Direction
///int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

/**************************************************************************************
* ///////////////////////////////////////////////////////////////////////////////////*
*************************************************************************************/

///==========CONSTANTS==============///
/// Digit 0123456789*#@%&^$"- ///
#define MX 100005
#define MXX 12500005
#define inf 1000000000
#define MD 1000000007LL
#define eps 1e-9
///================================///

int par[28];
int ind[28];
int oud[28];

int GP(int x) {
if( par[x]==x ) return x;
return (par[x]=GP( par[x] ));
}

void MU(int x,int y) {
int xx=GP(x),yy=GP(y);
if( xx!=yy ) {
par[ xx ]=yy;
}
}

int main() {
int tc, cs = 1, i, j, k, x, y,n,m;
S(tc);
while(tc--) {
S(n);
CLR(ind);
CLR(oud);
fr(i,0,25)par[i]=i;
fr(i,1,n) {
string s;
cin>>s;
int ln=SZ(s);
ind[ s[0]-'a' ]++;
oud[ s[ln-1]-'a' ]++;
MU( s[0]-'a' ,s[ln-1]-'a' );
}
int tot=0,a=0,b=0,f=1;
set<int>St;
fr(i,0,25) if( ind[i] || oud[i] )St.insert( GP(i) );
if( SZ(St) >1 )f=0;
fr(i,0,25) {
if(!f)break;
if( !ind[i] && !oud[i])continue;
if( abs(ind[i]-oud[i])>=2 ) {
f=0;
break;
}
if( ind[i]>oud[i] )a++;
if( ind[i]<oud[i] )b++;
}
if( f && (a+b==0 || (a==1&&b==1)) ) printf("Ordering is possible.\n");
else printf("The door cannot be opened.\n");

}
return 0;
}

Comments

Popular posts from this blog

HackerEarth: City and Campers 2 (DSU-UnionFind)

Two Pointers Technique & Binary Search For Beginners

Problem : Codeforces Round #406 (Div. 1) B [ Legacy ]( Dijakstra, Segment tree)