HDU 3518 Boring counting

后缀自动机。。

Boring counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1826    Accepted Submission(s): 733

Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

 

Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).

 

Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.

 

Sample Input

   
   
   
   

    
    
    
    aaaa
ababcabb
aaaaaa
#
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
3
3
   
   
   
   

 

Source

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU

 

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zhengfeng

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define ll long long
#define prt(k) cout<<#k"="<<k<<" ";

/**Suffix Automaton*/
const int Char=26,N=600022;
struct SAM_Node
{
    SAM_Node *fa,*ch[Char];
    int len,id,pos;
    SAM_Node() {}
    SAM_Node(int _len)
    {
        fa=0; len=_len;
        memset(ch,0,sizeof ch);
    }
}pool[N],*root,*last,*tail;   ///tail
int SAM_size;
#define sam pool
#define node SAM_Node
SAM_Node *newnode(int len)
{
    pool[SAM_size]=node(len);
    pool[SAM_size].id=SAM_size;
    return &pool[SAM_size++];
}
node *newnode(node *p)
{
    pool[SAM_size]=*p;
    pool[SAM_size].id=SAM_size;
    return &pool[SAM_size++];
}
void SAM_init()
{
    SAM_size=0;
    root=last=newnode(0);
    pool[0].pos=0;
}
void add(int x,int len)
{
    node *p=last,*np=newnode(p->len+1);
    np->pos=len; last=np;
    for(;p&&!p->ch[x];p=p->fa) p->ch[x]=np;
    if(!p) { np->fa=root;return ; }
    node *q=p->ch[x];
    if(q->len==p->len+1) { np->fa=q;return; }
    node *nq=newnode(q);
    nq->len=p->len+1;
    q->fa=nq; np->fa=nq;
    for(;p&&p->ch[x]==q;p=p->fa) p->ch[x]=nq;
}
void SAM_build(char *s)
{
    SAM_init();
    int len=strlen(s);
    for(int i=0;i<len;i++) add(s[i]-'a',i+1);
}
void Max(int &a,int b) { a=max(a,b); }
void Min(int &a,int b) { a=min(a,b); }
int c[N]; int L[N],R[N],num[N];
node *top[N];
char s[N];
#define inf 0x3f3f3f3f
void gao(int len)
{
    memset(c,0,sizeof c);
    memset(top,0,sizeof top);
    for(int i=0;i<SAM_size;i++) c[sam[i].len]++;
    for(int i=1;i<=len;i++) c[i]+=c[i-1];
    for(int i=0;i<SAM_size;i++) top[--c[sam[i].len]]=&sam[i];
    memset(L,63,sizeof L);
    memset(R,0,sizeof R);
    memset(num,0,sizeof num);
    for( node *p=root;;p=p->ch[s[p->len]-'a'])
    {
        num[p->id]=1;
        L[p->id]=R[p->id]=p->pos;
        if(p->len==len) break;
    }
    for(int i=SAM_size-1;i>=0;i--)
    {
        node *p=top[i];
        if(L[p->id]==inf&&R[p->id]==0)
        {
            L[p->id]=R[p->id]=p->pos;
        }
        if(p->fa) {
            node *q=p->fa;
            Min(L[q->id],L[p->id]);
            Max(R[q->id],R[p->id]);
            num[q->id]+=num[p->id];
        }
    }
    ll ans=0;
    for(int i=1;i<SAM_size;i++)
    {
        if(num[sam[i].id]<=1) continue;
        int ma=sam[i].len;
        int mi=sam[i].fa->len+1;
        int _len=R[sam[i].id]-L[sam[i].id];
        if(_len>=ma) ans+=ma-mi+1;
        else if(_len<ma&&_len>=mi) ans+=_len-mi+1;
    }
    printf("%I64d\n",ans);
}

int main()
{
    while(scanf("%s",s)==1&&s[0]!='#')
    {
        SAM_build(s);
        gao(strlen(s));
    }
}