题面

题解

我们把字符的出现次数哈希起来，然后把每个点向能在它之后的点连边。那么这显然是一个\(DAG\)，直接求最长路就行了

//minamoto#include
    
     #define R register#define fp(i,a,b) for(R int i=(a),I=(b)+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(char *s){    R int len=0;R char ch;while(((ch=getc())>'z'||ch<'a'))if(ch==EOF)return EOF;    for(s[++len]=ch;(ch=getc())>='a'&&ch<='z';s[++len]=ch);    return s[len+1]='\0',len;}char sr[1<<21],z[20];int C=-1,Z=0;inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}void print(char *s){    if(C>(1<<20))Ot();    for(R int i=1;s[i];++i)sr[++C]=s[i];    sr[++C]='\n';}const int N=10005,P=1e7,Base=233;struct eg{int v,nx;}e[N<<1];int head[N],deg[N],tot;inline void add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot,++deg[v];}char s[N][105];int q[N],len[N],ch[N][26],p[P],st[N],dis[N],Pre[N];int n,m,ans,pos,t,h,u,top;inline int get(int *s){    int res=0;    fp(i,0,25)res=(233ll*res+s[i])%P;    return res;}int main(){//  freopen("testdata.in","r",stdin);    while(~(++n,len[n]=read(s[n])));--n;    fp(i,1,n){        fp(j,1,len[i])++ch[i][s[i][j]-'a'];        p[get(ch[i])]=i;    }    fp(i,1,n)for(R int j=0,v;j<26;++j){        ++ch[i][j];        p[v=get(ch[i])]?(add(i,p[v]),0):0;        --ch[i][j];    }    h=1,t=0;    fp(i,1,n)!deg[i]?q[++t]=i,dis[i]=1:0;    while(h<=t){        u=q[h++];        go(u)(dis[v]=dis[u]+1,Pre[v]=u,!--deg[v])?q[++t]=v:0;    }    fp(i,1,n)dis[i]>ans?ans=dis[pos=i]:0;    while(pos)st[++top]=pos,pos=Pre[pos];    printf("%d\n",ans);    fd(i,top,1)print(s[st[i]]);    return Ot(),0;}