CF508D：Tanya and Password（欧拉通路 & 输出路径）

最新推荐文章于 2021-11-10 11:40:36 发布

junior19

最新推荐文章于 2021-11-10 11:40:36 发布

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分类专栏：欧拉路

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小女孩塔尼亚误删了父亲的秘密数据库密码，仅留下所有可能的连续三个字母的子串。通过构建哈希表并使用欧拉路径算法，本文探讨如何从这些子串中恢复原始密码。

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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.

Examples

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

题意：给N个长度为3的字符串，问能否组成总长度为N+2的字符串，如abcde=abc+bcd+cde，能就输出给字符串。

思路：每个字串后两位要跟另一个串的前两位一样才能接起来，那么将每个串的前两位和后两位拆开Hash后建边，跑一遍欧拉通路就行。

//referencr：UniqueColor
# include <bits/stdc++.h>
# define pb push_back
using namespace std;
const int maxn = 2e5+30;
vector<int>edge[maxn];
int cnt[maxn], in[maxn], out[maxn];
string ans;
void dfs(int x)
{
    while(cnt[x] < edge[x].size())
        dfs(edge[x][cnt[x]++]);
    ans += (char)(x%256);
}
int main()
{
    int n, u, v;
    char s[5];
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%s",s);
        u = s[0]*256+s[1];
        v = s[1]*256+s[2];
        edge[u].pb(v);
        ++out[u];
        ++in[v];
    }
    int _s=0, _e=0, st=u;//有可能起点和终点的入出度差均为0。
    for(int i=0; i<maxn; ++i)
    {
        int sub = in[i] - out[i];
        if(sub == -1) ++_s, st=i;
        else if(sub == 1) ++_e;
        else if(sub != 0)return 0*puts("NO");
    }
    if(_s>1 || _e>1 || _s + _e == 1) return 0*puts("NO");
    dfs(st);
    ans += (char)(st/256);
    if(ans.size() != n+2) puts("NO");
    else
    {
        puts("YES");
        reverse(ans.begin(), ans.end());
        cout << ans << endl;
    }
    return 0;
}