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 ofn+2 characters. She has written all the possiblen 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
5acaabaabacabbac
Output
YESabacaba
Input
4abcbCbcb1b13
Output
NO
Input
7aaaaaaaaaaaaaaaaaaaaa
Output
YESaaaaaaaaa
哈希一次字符串，，然后跑一次欧拉路
#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;map<string,int>toint;vector<string>tostring;vector<int>box;vector<int>edge[62*62*62+10];void seek(int from){while(1){box.push_back(from);if(edge[from].size()){int to=edge[from].back();edge[from].pop_back();from=to;}elsebreak;}}vector<int>ans;void work(int s){box.push_back(s);while(box.size()){int from=box.back();box.pop_back();if(edge[from].size())seek(from);elseans.push_back(from);}}int father[62*62*62+10];int run(int from){return father[from]==from?from:father[from]=run(father[from]);}bool isok(){int sum=0;for(int i=0;i<toint.size()&&sum<2;i++)if(father[i]==i)sum++;return sum<=1;}vector<string>in;int ind[62*62*62+10],out[62*62*62+10];int main(){int n;cin>>n;while(n–){string s;cin>>s;in.push_back(s.substr(0,2));in.push_back(s.substr(1,2));}for(int i=0;i<in.size();i++)toint[in[i]];map<string,int>::iterator it=toint.begin();for(int i=0;it!=toint.end();it++,i++){tostring.push_back(it->first);it->second=i;}for(int i=0;i<toint.size();i++)father[i]=i;for(int i=0;i<in.size();i+=2)father[run(toint[in[i]])]=run(toint[in[i+1]]);if(isok()){for(int i=0;i<in.size();i+=2){out[toint[in[i]]]++;ind[toint[in[i+1]]]++;edge[toint[in[i]]].push_back(toint[in[i+1]]);}int s=0,sum=0;for(int i=0;i<toint.size()&&sum<3;i++){if(ind[i]!=out[i]){if(abs(ind[i]-out[i])==1){sum++;if(ind[i]<out[i])s=i;}elsesum+=3;}}if(sum>2)cout<<"NO";else{cout<<"YES"<<endl;work(s);cout<<tostring[ans.back()];for(int i=ans.size()-2;i>-1;i–)cout<<tostring[ans[i]][1];}}elsecout<<"NO";}
不会因为别人显赫的成功而促使自己有卓越的进步。
