SPOJ8222(后缀自动机--出现次数最多的子串)

题目：http://www.spoj.com/problems/NSUBSTR/

 

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))

 

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>using namespace std;const int N=250005;struct State{    State *pre,*go[26];    int step,v;    void clear()    {        v=0;        pre=0;        step=0;        memset(go,0,sizeof(go));    }}*root,*last;State statePool[N*2],*b[2*N],*cur;void init(){    cur=statePool;    root=last=cur++;    root->clear();}void Insert(int w){    State *p=last;    State *np=cur++;    np->clear();    np->step=p->step+1;    while(p&&!p->go[w])        p->go[w]=np,p=p->pre;    if(p==0)        np->pre=root;    else    {        State *q=p->go[w];        if(p->step+1==q->step)            np->pre=q;        else        {            State *nq=cur++;            nq->clear();            memcpy(nq->go,q->go,sizeof(q->go));            nq->step=p->step+1;            nq->pre=q->pre;            q->pre=nq;            np->pre=nq;            while(p&&p->go[w]==q)                p->go[w]=nq, p=p->pre;        }    }    last=np;}char str[N];int cnt[N];int dp[N];int main(){    int n,m;    while(~scanf("%s",str))    {        n=strlen(str);        init();        for(int i=0; i<n; i++)            Insert(str[i]-'a');        memset(cnt,0,sizeof(cnt));        for(State *p=statePool; p!=cur; p++)            cnt[p->step]++;        for(int i=1; i<=n; i++)            cnt[i]+=cnt[i-1];        for(State *p=statePool; p!=cur; p++)            b[--cnt[p->step]]=p;        for(int i=0;i<n;i++)        {            int x=str[i]-'a';            root=root->go[x];            root->v++;        }        memset(dp,0,sizeof(dp));        int num=cur-statePool;        for(int i=num-1;i>0;i--)        {            dp[b[i]->step]=max(dp[b[i]->step],b[i]->v);            if(b[i]->pre) b[i]->pre->v += b[i]->v;        }        for(int i=n-1;i>0;i--)            dp[i]=max(dp[i],dp[i+1]);        for(int i=1;i<=n;i++)            printf("%d\n",dp[i]);    }    return 0;}