SPOJ 8222 [后缀自动机]

题目

NSUBSTR - Substrings

You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string ‘ababa’ F(3) will be 2 because there is a string ‘aba’ that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

输入

String S consists of at most 250000 lowercase latin letters.

输出

Output |S| lines. On the i-th line output F(i).

样例输入

ababa

样例输出

3
2
2
1
1

分析

没什么好分析的hh,基本裸的后缀自动机,对parent树进行Tsort(),自底向上维护r[i]的大小,注意开始赋值时每个实节点(即np,而非nq)的r赋为1,其他为0

完整代码

#include<bits/stdc++.h>#define maxn 500010//#define DEBUGusing namespace std;int n;int sum[maxn],tmp[maxn],f[maxn];char ch[maxn];struct SAM{    int root,tot,last;    int son[maxn][26],fa[maxn],maxl[maxn],r[maxn];    int addnode(int n) {return maxl[++tot]=n,tot;}     void init()    {        root=tot=last=1;        memset(son,0,sizeof(son));        memset(fa,0,sizeof(fa));        memset(maxl,0,sizeof(maxl));        memset(r,0,sizeof(r));    }    void add(int pos)    {        int x=ch[pos]-'a',np=addnode(pos),p=last;        last=np,r[np]=1;        for( ; p&&!son[p][x] ; p=fa[p] ) son[p][x]=np;        if(!p) fa[np]=root;        else        {            int q=son[p][x];            if(maxl[q]==maxl[p]+1) fa[np]=q;            else            {                int nq=addnode(maxl[p]+1);                memcpy(son[nq],son[q],sizeof(son[q]));                fa[nq]=fa[q];                fa[np]=fa[q]=nq;                for( ; son[p][x]==q ; p=fa[p] ) son[p][x]=nq;            }        }    }    void Tsort()    {        for(int i=1;i<=tot;++i) sum[maxl[i]]++;        for(int i=1;i<=n;++i) sum[i]+=sum[i-1];        for(int i=1;i<=tot;++i) tmp[sum[maxl[i]]--]=i;    #ifdef DEBUG        for(int i=1;i<=tot;++i) printf("   tmp[%d]=%d   maxl[tmp[%d]]=%d\n",i,tmp[i],i,maxl[tmp[i]]);    #endif    }    void build()    {        init();        scanf("%s",ch+1);        n=strlen(ch+1);        for(int i=1;i<=n;++i) add(i);        Tsort();    }    void work()    {        build();    #ifdef DEBUG        for(int i=1;i<=tot;++i)            cerr<<"r["<<i<<"]="<<r[i]<<endl;    #endif        for(int i=tot;i;i--) r[fa[tmp[i]]]+=r[tmp[i]];        for(int i=1;i<=tot;++i) f[maxl[i]]=max(f[maxl[i]],r[i]);        for(int i=1;i<=n;++i) printf("%d\n",f[i]);    }} sam ;int main(){    sam.work();    return 0;}/*Input:ababanoiOutput:32211*/