CF 508D(Tanya and Password-欧拉路径，弗罗莱算法)

D. Tanya and Password

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters.
She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password
out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.

Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted
of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.

Input

The first line contains integer n (1 ≤ n ≤ 2·105),
the number of three-letter substrings Tanya got.

Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or
a digit.

Output

If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".

If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.

Sample test(s)

input

5
aca
aba
aba
cab
bac

output

YES
abacaba

input

4
abc
bCb
cb1
b13

output

NO

input

7
aaa
aaa
aaa
aaa
aaa
aaa
aaa

output

YES
aaaaaaaaa

欧拉路径，弗罗莱算法 主要参见下面的资料：

/article/8868248.html

本题给出一个字符串所有的连续3位的字符串，要求找出一个满足条件的原字符串。

把2个字符(如‘aa')为点，

每个3字符串，前2字符向后两字符连边(如'abc'：'ab'--->‘bc')

原题转为求新图上的欧拉路径。

PS:注意特判图不连通的情况,注意删边。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXLen (200000+10)
#define MAXN (200000+10)
#define MAXM (MAXLen)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,N;
char s[4];
int pre[MAXN]={0},next[MAXM]={0},edge[MAXM],size=0;
void addedge(int u,int v)
{
edge[++size]=v;
next[size]=pre[u];
pre[u]=size;
}
int indegree[MAXN],outdegree[MAXN];
int h(char c)
{
if ('a'<=c&&c<='z') return c-'a'+1;
if ('A'<=c&&c<='Z') return c-'A'+27;
else return c-'0'+1+26+26;
}
int h(char c1,char c2)
{
return h(c1)*63+h(c2);
}
int ans[MAXLen],anstot=0,S[MAXN],tot=0;
int hash[MAXLen];
void dfs(int x)
{
Forp(x)
{
int v=edge[p];
pre[x]=next[p];
S[++tot]=v;
dfs(v);
return ;
}
}
void Fleury(int st)
{
S[++tot]=st;
while (tot)
{
int x=S[tot];
if (pre[x]) dfs(x);
else
{
ans[++anstot]=x;
tot--;
}
}

if (anstot^(n+1))
{
cout<<"NO\n";
return;
}
cout<<"YES\n";
ForD(i,anstot)
{
printf("%c",hash[ans[i]/63]);
}
printf("%c\n",hash[ans[1]%63]);

}
int main()
{
//	freopen("CF508D.in","r",stdin);
//	freopen(".out","w",stdout);

Fork(i,'a','z') hash[i-'a'+1]=i;
Fork(i,'A','Z') hash[i-'A'+27]=i;
Fork(i,'0','9') hash[i-'0'+1+26+26]=i;

cin>>n;
int st=0;
For(i,n)
{
scanf("%s",s);
addedge(h(s[0],s[1]),h(s[1],s[2]));
outdegree[h(s[0],s[1])]++,indegree[h(s[1],s[2])]++;
st=h(s[0],s[1]);
}
N=h('9','9');
int s1=0,s2=0,s3=0;
For(i,N)
{
int t=indegree[i]-outdegree[i];
if (t==1) ++s1;
else if (t==0) ++s2;
else if (t==-1) ++s3,st=i;
else
{
cout<<"NO\n";
return 0;
}
}

if ((s1==s3&&s3==0)||(s1==s3&&s3==1))
{
Fleury(st);
return 0;
}

cout<<"NO\n";
return 0;

return 0;
}