CodeForces 508D Tanya and Password(欧拉路径)

题目：​​http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=111980​​

​​Tanya and Password​​

​​Submit​​​ ​​Status​​

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

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

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

这是有向图的欧拉通路问题，虽然我能看出来，但是输出问题和存储问题没有很好解决。参考了大神们的代码：发现他们用0,1字符和1,2字符作为线段端点的数字信息。妙！输出则是DFS。

#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;
}