NOI2018 your name (68pts)

NOI2018 your name (68pts)

The meaning of problems

To a string \ (S \) , to give \ (m \) string \ (T \) , Q \ (T \) the number of continuous non-empty string is not \ (S \) of consecutive non-null words string .
\ [| S | \ le 5 \ times10 ^ 5, \ sum {T} \ le10 ^ 6 \]

answer

We can put the title into \ (T \) has a different number of sub-string is not in the nature of \ (S \) appeared, we have built two strings \ (SAM \) , first obtains \ (T \) each prefix \ (S \) a maximum matching length \ (L_i \) (with \ (S \) a \ (the SAM \) find the longest common prefix can be determined \ (L_i \) ), and then in \ (T \) is \ (SAM \) seek the answer. We define nodes \ (I \) represents a collection of \ (T \) position first appears as \ (pos_i \) , then
\ [ans = \ sum_ {i = 1} ^ {tot} {max (0 , len [i] -max (len [link [i]], l_ {pos_i}))} \]

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
int read()
{
    int k=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) k=k*10+c-'0';return k*f;
}
const int N=2000055;
int n,m,tot,last,link[N],len[N],ch[N][26];
int Len[N];
char s[N],t[N];
void init()
{
    tot=last=0;
    link[0]=-1;len[0]=0;
}
void extend(int c)
{
    int p=last,cur=++tot;
    len[cur]=len[p]+1;
    for(;p!=-1&&!ch[p][c];p=link[p]) ch[p][c]=cur;
    if(p==-1) link[cur]=0;
    else 
    {
        int q=ch[p][c];
        if(len[p]+1==len[q]) link[cur]=q;
        else 
        {
            int nq=++tot;
            memcpy(ch[nq],ch[q],sizeof(ch[q]));
            len[nq]=len[p]+1;link[nq]=link[q];
            link[q]=link[cur]=nq;
            for(;p!=-1&&ch[p][c]==q;p=link[p]) 
                ch[p][c]=nq;
        }
    }
    last=cur;
}
struct sam
{       
    int tot=0,last=0,link[N],len[N],ch[N][26],pos[N];
    void init()
    {
        for(int i=0;i<=tot;i++)
            pos[i]=link[i]=len[i]=0,memset(ch[i],0,sizeof(ch[i]));
        tot=last=0;
        link[0]=-1;len[0]=0;
    }
    void extend(int c,int x)
    {
        int p=last,cur=++tot;
        len[cur]=len[p]+1;pos[cur]=x;
        for(;p!=-1&&!ch[p][c];p=link[p]) ch[p][c]=cur;
        if(p==-1) link[cur]=0;
        else 
        {
            int q=ch[p][c];
            if(len[p]+1==len[q]) link[cur]=q;
            else 
            {
                int nq=++tot;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                len[nq]=len[p]+1;link[nq]=link[q];pos[nq]=pos[q];
                link[q]=link[cur]=nq;
                for(;p!=-1&&ch[p][c]==q;p=link[p]) 
                    ch[p][c]=nq;
            }
        }
        last=cur;
    }
    ll query()
    {
        ll ans=0;
        for(int i=1;i<=tot;i++)
            ans+=max(0,len[i]-max(len[link[i]],Len[pos[i]]));;
        return ans;
    }
}sm;
int main()
{
    int a,b;
    scanf("%s",s);
    n=strlen(s);
    init();
    for(int i=0;i<n;i++)
        extend(s[i]-'a');
    m=read();
    for(int i=1;i<=m;i++)
    {
        scanf("%s",t);a=read();b=read();
        n=strlen(t);
        sm.init();
        int now=0,l=0;ll ans=0;
        for(int j=0;j<n;j++)
        {
            sm.extend(t[j]-'a',j);
            while(now&&!ch[now][t[j]-'a']) now=link[now],l=len[now];
            if(ch[now][t[j]-'a']) l++,now=ch[now][t[j]-'a'];
            Len[j]=l;
        }
        ans=sm.query();
        printf("%lld\n",ans);
    }
    return 0;
}