题意：
给一个字符串S，令F(x)表示S的所有长度为x的子串中，出现次数的最大值。
求F(1)..F(Length(S)) Length(S) <= 250000
思路：板子中st[x]定义为root到x的最多步数，可以用来更新所有长度为[1..st[x]]的答案
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 typedef pair<ll,int>P; 11 #define N 510000 12 #define M 151000 13 #define fi first 14 #define se second 15 #define MP make_pair 16 #define pi acos(-1) 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 19 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 20 #define lowbit(x) x&(-x) 21 #define Rand (rand()*(1<<16)+rand()) 22 #define id(x) ((x)<=B?(x):m-n/(x)+1) 23 #define ls p<<1 24 #define rs p<<1|1 25 26 const int MOD=998244353,inv2=(MOD+1)/2; 27 double eps=1e-6; 28 ll INF=1e18; 29 ll inf=5e13; 30 int dx[4]={-1,1,0,0}; 31 int dy[4]={0,0,-1,1}; 32 33 char ch[N]; 34 int n,i,x,p,q,np,nq,cnt,L,st[N],c[N][26],f[N],pos[N],bl[N],to[N],b[N],sz[N],ans[N]; 35 36 int read() 37 { 38 int v=0,f=1; 39 char c=getchar(); 40 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 41 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 42 return v*f; 43 } 44 45 void add(int x) 46 { 47 p=np; 48 st[np=++cnt]=st[p]+1; 49 to[np]=i; 50 pos[i]=np; 51 while(p&&!c[p][x]) 52 { 53 c[p][x]=np; 54 p=f[p]; 55 } 56 if(!p) f[np]=1; 57 else if(st[p]+1==st[q=c[p][x]]) f[np]=q; 58 else 59 { 60 st[nq=++cnt]=st[p]+1; 61 memcpy(c[nq],c[q],sizeof c[q]); 62 f[nq]=f[q]; 63 f[q]=f[np]=nq; 64 while(p&&c[p][x]==q) 65 { 66 c[p][x]=nq; 67 p=f[p]; 68 } 69 } 70 } 71 72 73 int main() 74 { 75 //freopen("1.in","r",stdin); 76 //freopen("1.out","w",stdout); 77 np=cnt=1; 78 scanf("%s",ch); 79 n=strlen(ch); 80 rep(i,0,n-1) add(ch[i]-'a'); 81 rep(i,1,cnt) b[st[i]]++; 82 rep(i,1,n) b[i]+=b[i-1]; 83 rep(i,1,cnt) bl[b[st[i]]--]=i; 84 for(i=0,p=1;i<n;i++) sz[p=c[p][ch[i]-'a']]++; 85 for(i=cnt;i;i--) sz[f[bl[i]]]+=sz[bl[i]]; 86 //rep(i,0,n-1) printf("%d ",sz[pos[i]]); 87 rep(i,1,n) ans[i]=0; 88 rep(i,1,cnt) ans[st[i]]=max(ans[st[i]],sz[i]); 89 per(i,n-1,1) ans[i]=max(ans[i],ans[i+1]); 90 rep(i,1,n) printf("%d ",ans[i]); 91 return 0; 92 }