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：
http://blog.csdn.net/u012659423/article/details/43319245
#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;
}