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-6;
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=0;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) {x=x*10+c-'0'; c=getchar();}
}

const int maxn=100100;
char s[maxn];
int len,f[maxn][2];
vector<string>ans;
map<string,bool>d;

int main()
{
    memset(s,0,sizeof s);
    scanf("%s",s); len=strlen(s);
    f[len][0]=f[len][1]=1;
    for(int i=len-1;i>4;i--)
    {
        if(len-i>=2)
        {
            if(f[i+2][1]==1)
            {
                char t[5]; memset(t,0,sizeof t);
                t[0]=s[i]; t[1]=s[i+1];

                f[i][0]=1;
                if(d[t]==0) ans.push_back(t),d[t]=1;
            }
            else if(f[i+2][0]==1)
            {
                char t[5],g[5]; memset(t,0,sizeof t); memset(g,0,sizeof g);
                t[0]=s[i]; t[1]=s[i+1]; g[0]=s[i+2]; g[1]=s[i+3];

                if(strcmp(t,g)==0) continue;
                f[i][0]=1;
                if(d[t]==0) ans.push_back(t),d[t]=1;
            }
        }
        if(len-i>=3)
        {
            if(f[i+3][0]==1)
            {
                char t[5]; memset(t,0,sizeof t);
                t[0]=s[i]; t[1]=s[i+1]; t[2]=s[i+2];

                f[i][1]=1;
                if(d[t]==0) ans.push_back(t),d[t]=1;
            }

            else if(f[i+3][1]==1)
            {
                char t[5],g[5]; memset(t,0,sizeof t); memset(g,0,sizeof g);
                t[0]=s[i]; t[1]=s[i+1]; t[2]=s[i+2];
                g[0]=s[i+3]; g[1]=s[i+4]; g[2]=s[i+5];

                if(strcmp(t,g)==0) continue;
                f[i][1]=1;
                if(d[t]==0) ans.push_back(t),d[t]=1;
            }
        }
    }
    sort(ans.begin(),ans.end());
    printf("%d\n",ans.size());
    for(int i=0;i<ans.size();i++) cout<<ans[i]<<endl;
    return 0;
}