Codeforces Round #288 Div2 D

Problem

给N个三元字符串 Si ，每个字符串只含有3个字符。现将N个字符串任意排序，使得对于 i∈[1,N−1] 满足 Si[1..2]=Si+1[0..1] 成立（比如：aba,bab,abc,bce）。问是否可以成立，若可以输出合并后的字符串（比如：aba,bab,abc,bce->ababce）

Limits

TimeLimit(ms):2000

MemoryLimit(MB):256

N∈[1,2×105]

字符集∈[a,z]⋃[A,Z]⋃[0,9]

Look up Original Problem From here

Solution

用有向图的欧拉路径来处理。

1. 建图：对于每个三元字符串 Si ，令hash1= Si[0..1] ，hash2= Si[1..2] ，将hash1->hash2连一条有向边，

2. 判断图G是否存在欧拉路径：

• G是一个联通图
• G中所有点入度＝出度，或者，G中存在一个点入度－出度＝1，一个点出度－入度＝1，其余点入度＝出度

3. 若存在，dfs求出答案

Complexity

TimeComplexity:O(N)

MemoryComplexity:O(N)

My Code

//Hello. I'm Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-6
#define MOD 1000000007
#define MAXN 5625
#define N
#define M
int idx(char c){
    return c-'0';
}
const int hashval=75;
int strhash(char c1,char c2){
    return idx(c1)*hashval+idx(c2);
}
int n,maxi,origin;
queue<int>node[MAXN];
char s[10];
int indegree[MAXN],outdegree[MAXN];
string str[MAXN];
string make_string(char c1,char c2){
    string res;
    res.clear();
    res+=c1,res+=c2;
    return res;
}
void printno(){
    printf("NO\n");
    exit(0);
}
stack<int>st;
void dfs(int now){
    while(!node[now].empty()){
        int to=node[now].front();
        node[now].pop();
        dfs(to);
    }
    st.push(now);
}
bool vis[MAXN];
int main(){
    scanf("%d",&n);
    maxi=0;
    origin=INT;
    repin(i,1,n){
        scanf("%s",s);
        int v1=strhash(s[0],s[1]);
        int v2=strhash(s[1],s[2]);
        node[v1].push(v2);
        maxi=max(maxi,v1),maxi=max(maxi,v2);
        if(i==1) origin=v1;
        outdegree[v1]+=1,indegree[v2]+=1;
        if(!vis[v1]) str[v1]=make_string(s[0],s[1]),vis[v1]=true;
        if(!vis[v2]) str[v2]=make_string(s[1],s[2]),vis[v2]=true;
    }
    int num01=0,num10=0,numodd=0;
    repin(i,0,maxi){
        if(indegree[i]!=outdegree[i]){
            numodd+=1;
            if(indegree[i]-outdegree[i]==1) num10+=1;
            else if(outdegree[i]-indegree[i]==1) num01+=1,origin=i;
            else printno();
        }
    }
    if(numodd!=0 && numodd!=2) printno();
    else if(numodd==2 && (num10!=1 || num01!=1)) printno();
    dfs(origin);
    repin(i,0,maxi){//判断图是否整体连通，这一步容易被忽略..
        if(!node[i].empty()) printno();
    }
    printf("YES\n");
    string ans;
    while(!st.empty()){
        int now=st.top();
        st.pop();
        ans+=str[now][0];
        if(st.empty())  ans+=str[now][1];
    }
    cout<<ans<<endl;
}