Codeforces Round #349 (Div. 2) C. Reberland Linguistics 【DP】

/* ***********************************************Author        :MaltubEmail         :[e-mail]          :htttp://www.xiang578.com************************************************ */#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>//#include <bits/stdc++.h>#define rep(i,a,n) for(int i=a;i<n;i++)#define per(i,a,n) for(int i=n-1;i>=a;i--)#define pb push_backusing namespace std;typedef vector<int> VI;typedef long long ll;const ll mod=1000000007;const int N=2048;char s[100000+10];int dp[10000+10][5];map<string,int>mp;vector<string>v;int len;int get(int st,int l){    string t1,t2,t3;    if(st+l>len)        return 0;    else if(st+l==len)    {        for(int i=0; i<l; i++)            t1+=s[st+i];        if(mp[t1]==0)        {            mp[t1]++;            v.push_back(t1);        }    }    else    {        int f=0,s2=st+l;        for(int i=0; i<l; i++)            t1+=s[st+i];        for(int i=0;i<2;i++)            t2+=s[s2+i];        for(int i=0;i<3;i++)            t3+=s[s2+i];       if(l==2)       {           if(t1!=t2&&dp[s2][2]) f=1;           if(dp[s2][3]) f=1;       }       else if(l==3)       {           if(t1!=t3&&dp[s2][3]) f=1;           if(dp[s2][2]) f=1;       }        if(f==0) return 0;        if(mp[t1]==0)        {            mp[t1]++;            v.push_back(t1);        }    }    return 1;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    memset(s,0,sizeof(s));    scanf("%s",s);    v.clear();    mp.clear();    memset(dp,0,sizeof(dp));    len=strlen(s);    for(int i=len-1; i>=5; i--)    {        dp[i][2]=get(i,2);        dp[i][3]=get(i,3);    }    sort(v.begin(),v.end());    printf("%d\n",v.size());    for(int i=0; i<v.size(); i++)    {        cout<<v[i]<<endl;    }    return 0;}