CodeForces 667C Reberland Linguistics

$dp$。

题意中有一个词组：$in$ $a$ $row$，是连续的意思....

因此这题只要倒着$dp$一下就可以了。$f[i][0]$表示从$i$位置往后割两个能否割，$f[i][1]$表示从$i$位置往后割三个能否割。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c=getchar(); x=;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) {x=x*+c-''; c=getchar();}
}

const int maxn=;
char s[maxn];
int len,f[maxn][];
vector<string>ans;
map<string,bool>d;

int main()
{
    memset(s,,sizeof s);
    scanf("%s",s); len=strlen(s);
    f[len][]=f[len][]=;
    for(int i=len-;i>;i--)
    {
        if(len-i>=)
        {
            if(f[i+][]==)
            {
                char t[]; memset(t,,sizeof t);
                t[]=s[i]; t[]=s[i+];

                f[i][]=;
                if(d[t]==) ans.push_back(t),d[t]=;
            }
            else if(f[i+][]==)
            {
                char t[],g[]; memset(t,,sizeof t); memset(g,,sizeof g);
                t[]=s[i]; t[]=s[i+]; g[]=s[i+]; g[]=s[i+];

                if(strcmp(t,g)==) continue;
                f[i][]=;
                if(d[t]==) ans.push_back(t),d[t]=;
            }
        }
        if(len-i>=)
        {
            if(f[i+][]==)
            {
                char t[]; memset(t,,sizeof t);
                t[]=s[i]; t[]=s[i+]; t[]=s[i+];

                f[i][]=;
                if(d[t]==) ans.push_back(t),d[t]=;
            }

            else if(f[i+][]==)
            {
                char t[],g[]; memset(t,,sizeof t); memset(g,,sizeof g);
                t[]=s[i]; t[]=s[i+]; t[]=s[i+];
                g[]=s[i+]; g[]=s[i+]; g[]=s[i+];

                if(strcmp(t,g)==) continue;
                f[i][]=;
                if(d[t]==) ans.push_back(t),d[t]=;
            }
        }
    }
    sort(ans.begin(),ans.end());
    printf("%d\n",ans.size());
    for(int i=;i<ans.size();i++) cout<<ans[i]<<endl;
    return ;
}