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SPOJ 8222 NSUBSTR Substrings

2024-07-10 16:37:41   229人阅读


SAM的简单应用....

由SAM可知从root到达的每个节点所经过的路径都对着应原串的一个子串，每个节点能到几次接收态就等于这个子串出现了几次。从最后一个节点往上走，就可以用DP更新出每个子串出现了多少次。

出现了5次的子串一定也出现了4，3，2，1次。。。所以最后再用长度长的给长度小的更新一下。。。。

Substrings
Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

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|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

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

Example

Input:
ababa

Output:
3
2
2
1
1

Source

Immagination

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=600000;

struct SAM_Node
{
    SAM_Node *fa,*next[26];
    int len,id,pos;
    SAM_Node(){}
    SAM_Node(int _len)
    {
        len=_len; fa=0;
        memset(next,0,sizeof(next));
    }
};

SAM_Node SAM_node[maxn*2],*SAM_root,*SAM_last;
int SAM_size;

SAM_Node *newSAM_Node(int len)
{
    SAM_node[SAM_size]=SAM_Node(len);
    SAM_node[SAM_size].id=SAM_size;
    return &SAM_node[SAM_size++];
}

SAM_Node *newSAM_Node(SAM_Node *p)
{
    SAM_node[SAM_size]=*p;
    SAM_node[SAM_size].id=SAM_size;
    return &SAM_node[SAM_size++];
}

void SAM_init()
{
    SAM_size=0;
    SAM_root=SAM_last=newSAM_Node(0);
    SAM_node[0].pos=0;
}

void SAM_add(int x,int len)
{
    SAM_Node *p=SAM_last,*np=newSAM_Node(p->len+1);
    np->pos=len; SAM_last=np;
    for(;p&&!p->next[x];p=p->fa)
        p->next[x]=np;
    if(!p)
    {
        np->fa=SAM_root;
        return ;
    }
    SAM_Node *q=p->next[x];
    if(q->len==p->len+1)
    {
        np->fa=q;
        return ;
    }
    SAM_Node *nq=newSAM_Node(q);
    nq->len=p->len+1;
    q->fa=nq; np->fa=nq;
    for(;p&&p->next[x]==q;p=p->fa)
        p->next[x]=nq;
}

char str[maxn];
bool vis[maxn];
int r[maxn],dp[maxn];

int Find(SAM_Node *p)
{
    if(vis[p->id]==true)
        return r[p->id];
    vis[p->id]=true;
    r[p->id]=0;
    for(int i=0;i<26;i++)
    {
        if(p->next[i])
            r[p->id]+=Find(p->next[i]);
    }
    return r[p->id];
}

void get_count()
{
    memset(r,0,sizeof(r));
    memset(vis,0,sizeof(vis));
    SAM_Node *p=SAM_last;
    vis[p->id]=true;
    for(;p;p=p->fa)
    {
        Find(p);
        r[p->id]++;
    }
}

int main()
{
while(scanf("%s",str)!=EOF)
{
    int n=strlen(str);
    SAM_init();
    for(int i=0;i<n;i++)
        SAM_add(str[i]-'a',i+1);
    get_count();
    memset(dp,0,sizeof(dp));
    for(int i=0;i<SAM_size;i++)
    {
        dp[SAM_node[i].len]=max(dp[SAM_node[i].len],r[SAM_node[i].id]);
    }
    for(int i=n-1;i>=1;i--)
    {
        dp[i]=max(dp[i],dp[i+1]);
    }
    for(int i=1;i<=n;i++)
    {
        printf("%d\n",dp[i]);
    }
}
    return 0;
}





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