那可是D题啊@。@ 做不上我心安了

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88120#problem/B

Description

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 Input

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


开始还理解错题意了，还以为重叠的只有开头的一个字母和结尾的一个字母==但是是两个 对于这两个字符的哈希用的也挺好

欧拉回路，百闻不如一见==还是得先判断存在与否，根据入度出度，这个代码处理的好巧妙

最最重要的是dfs写回路！！

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N1=2e5+5,N2=4e3; // (26*2+10)^2<4000
int f[N2],in[N2],out[N2],path[N1],e[N2][N2];
bool has[N2];
int len,st,n;
struct node{
    int u,v;
    char name[5];
}snode[N1];
int abs(int x){
    return x>0?x:(-x);
}
void init(){
    for(int i=0;i<N2;i++) f[i]=i;
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(path,0,sizeof(path));
    memset(e,0,sizeof(e));
    memset(has,0,sizeof(has));
}
int find(int x){
    if(x==f[x]) return x;
    return f[x]=find(f[x]);
}
int ctoi(char ch){
    if(ch<='9'&&ch>='0') return ch-'0';
    if(ch>='A'&&ch<='Z') return ch-'A'+10;
    return ch-'a'+36;
}
char itoc(int x){
    if(x<=9&&x>=0) return x+'0';
    if(x<=35&&x>=10) return x+'A'-10;
    return x+'a'-36;
}
bool Exist(){
    int t=-1,sum=0,temp;
    for(int i=0;i<N2;i++){
        if(has[i]){
           if(t==-1) t=find(i);
           else if(find(i)!=t)  return 0;
        }
    }
    for(int i=0;i<N2;i++){
        if(has[i]){
             temp=i;
             if(in[i]!=out[i]){
                 sum++;
                 if(abs(in[i]-out[i])>1) return 0;
                 if(out[i]>in[i]) st=i; //链的端点
             }
        }
    }
    if(sum>2) return 0;
    if(sum==0) st=temp;  // 环的一节点
    return 1;
}
void dfs(int q){
    for(int i=0;i<N2;i++){
        while(e[q][i]){
            e[q][i]--;
            dfs(i);
            path[len++]=i;
        }
    }
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int n;
    while(cin>>n){
        init();
        for(int i=0;i<n;i++){
            scanf("%s",snode[i].name);
            snode[i].u=ctoi(snode[i].name[0])*62+ctoi(snode[i].name[1]);
            snode[i].v=ctoi(snode[i].name[1])*62+ctoi(snode[i].name[2]);
            int a=snode[i].u,b=snode[i].v;
            has[a]=has[b]=1;
            f[find(a)]=f[find(b)];
            out[a]++;
            in[b]++;
            e[a][b]++;
        }
        if(!Exist()) puts("NO");
        else {
             len=0;
             dfs(st);
             path[len++]=st;
             puts("YES");
             //cout<<path[len-1]<<endl;
             printf("%c%c",itoc(path[len-1]/62),itoc(path[len-1]%62));
             for(int i=len-2;i>=0;i--){
                 printf("%c",itoc(path[i]%62));
             }
             puts("");
        }
    }
    return 0;
}