SAM基本操作 拓扑求每个节点的  最左出现left，最右出现right，出现了几次num ......

对于每一个出现两次以上的节点，对其所对应的一串子串的长度范围 [fa->len+1,len] 和其最大间距 right-left比较

即可......

Boring counting

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#

2
3
3

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

using namespace std;

const int CHAR=26,maxn=1100;

struct SAM_Node
{
    SAM_Node *fa,*next[CHAR];
    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];
int len,c[maxn],L[maxn*2],R[maxn*2],num[maxn*2];
SAM_Node *top[maxn*2];

int main()
{
while(scanf("%s",str)!=EOF)
{
    if(str[0]=='#') break;
    SAM_init();
    len=strlen(str);
    for(int i=0;i<len;i++)
        SAM_add(str[i]-'a',i+1);

    memset(c,0,sizeof(c)); memset(top,0,sizeof(top));
    memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); memset(num,0,sizeof(num));

    ///get tupo sort
    for(int i=0;i<SAM_size;i++)
        c[SAM_node[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_node[i].len]]=&SAM_node[i];

    ///get L,R,num
    SAM_Node *p=SAM_root;
    for(;p->len!=len;p=p->next[str[p->len]-'a'])
    {
        num[p->id]=1;
        L[p->id]=R[p->id]=p->len;
    }
    for(int i=SAM_size-1;i>=0;i--)
    {
        p=top[i];
        if(L[p->id]==0&&R[p->id]==0)
        {
            L[p->id]=R[p->id]=p->pos;
        }
        if(p->fa)
        {
            SAM_Node *q=p->fa;
            num[q->id]+=num[p->id];
            if(L[q->id]==0||L[q->id]>L[p->id])
                L[q->id]=L[p->id];
            if(R[q->id]==0||R[q->id]<R[p->id])
                R[q->id]=R[p->id];
        }
    }
    int ans=0;
    for(int i=1;i<SAM_size;i++)
    {
        int ma=SAM_node[i].len;
        int mi=SAM_node[i].fa->len+1;
        int le=R[SAM_node[i].id]-L[SAM_node[i].id];
        if(le>=ma)
            ans+=ma-mi+1;
        else if(le>mi)
            ans+=le-mi+1;
    }
    printf("%d\n",ans);
}
    return 0;
}

HDOJ 3518 Boring counting,布布扣,bubuko.com

HDOJ 3518 Boring counting

原文：http://blog.csdn.net/ck_boss/article/details/36891749