HDU 4622 Reincarnation （区间不相同子串个数：字符串哈希 | 后缀数组 | 后缀自动机）

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Problem Description

Now you are back,and have a task to do:

Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.

And you have some query,each time you should calculate f(s[l…r]), s[l…r] means the sub-string of s start from l end at r.

Input

The first line contains integer T(1<=T<=5), denote the number of the test cases.

For each test cases,the first line contains a string s(1 <= length of s <= 2000).

Denote the length of s by n.

The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.

Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.

Output

For each test cases,for each query,print the answer in one line.

Sample Input

2bbaba53 42 22 52 41 4baaba53 33 41 43 55 5

Sample Output

3175813851

题意

求区间不相同子串个数。

思路

模板题目，有好多种解法：字符串哈希、后缀数组、后缀自动机…

AC 代码

字符串哈希

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<cmath>#include<algorithm>#include<iostream>using namespace std;const int HASH = 10007;const int MAXN = 2100;struct hashmap{    int head[HASH],next[MAXN],size;    unsigned long long state[MAXN];    int f[MAXN];    void init()    {        size=0;        memset(head,-1,sizeof(head));    }    int insert(unsigned long long val,int _id)    {        int h=val%HASH;        for(int i=head[h]; i!=-1; i=next[i])        {            if(val==state[i])            {                int temp=f[i];                f[i]=_id;                return temp;            }        }        f[size]=_id;        state[size]=val;        next[size]=head[h];        head[h]=size++;        return 0;    }} h;const int seed = 13331;unsigned long long P[MAXN];unsigned long long S[MAXN];char str[MAXN];int ans[MAXN][MAXN];int main(){    P[0]=1;    for(int i=1; i<MAXN; i++)        P[i]=P[i-1]*seed;    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s",str);        int n=strlen(str);        for(int i=1; i<=n; i++)            S[i]=S[i-1]*seed+str[i-1];        memset(ans,0,sizeof(ans));        for(int L=1; L<=n; L++)        {            h.init();            for(int i=1; i+L-1<=n; i++)            {                int l=h.insert(S[i+L-1]-S[i-1]*P[L],i);                ans[i][i+L-1]++;                ans[l][i+L-1]--;            }        }        for(int i=n; i>=0; i--)            for(int j=i; j<=n; j++)                ans[i][j]+=ans[i+1][j]+ans[i][j-1]-ans[i+1][j-1];        int Q;        scanf("%d",&Q);        while(Q--)        {            int a,b;            scanf("%d%d",&a,&b);            printf("%d\n",ans[a][b]);        }    }    return 0;}

后缀数组

#include<algorithm>#include<iostream>#include<string.h>#include<stdio.h>using namespace std;const int INF=0x3f3f3f3f;const int maxn=2010;typedef long long ll;char txt[maxn];int sa[maxn],T1[maxn],T2[maxn],ct[maxn],he[maxn],rk[maxn],n,m;void getsa(char *st){    int i,k,p,*x=T1,*y=T2;    for(i=0; i<m; i++) ct[i]=0;    for(i=0; i<n; i++) ct[x[i]=st[i]]++;    for(i=1; i<m; i++) ct[i]+=ct[i-1];    for(i=n-1; i>=0; i--)        sa[--ct[x[i]]]=i;    for(k=1,p=1; p<n; k<<=1,m=p)    {        for(p=0,i=n-k; i<n; i++) y[p++]=i;        for(i=0; i<n; i++) if(sa[i]>=k) y[p++]=sa[i]-k;        for(i=0; i<m; i++) ct[i]=0;        for(i=0; i<n; i++) ct[x[y[i]]]++;        for(i=1; i<m; i++) ct[i]+=ct[i-1];        for(i=n-1; i>=0; i--) sa[--ct[x[y[i]]]]=y[i];        for(swap(x,y),p=1,x[sa[0]]=0,i=1; i<n; i++)            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;    }}void gethe(char *st){    int i,j,k=0;    for(i=0; i<n; i++) rk[sa[i]]=i;    for(i=0; i<n-1; i++)    {        if(k) k--;        j=sa[rk[i]-1];        while(st[i+k]==st[j+k]) k++;        he[rk[i]]=k;    }}int main(){    int t,i,q,le,ri,lcp,ml,ans,tp;    scanf("%d",&t);    while(t--)    {        scanf("%s",txt);        n=strlen(txt)+1;        m=150;        getsa(txt);        gethe(txt);        scanf("%d",&q);        while(q--)        {            scanf("%d%d",&le,&ri);            le--,ri--;            ans=lcp=ml=0;            for(i=1; i<n; i++)            {                lcp=min(lcp,he[i]);                ml=min(ml,lcp);                if(sa[i]>=le&&sa[i]<=ri)                {                    ans+=ri-sa[i]+1-min(ri-sa[i]+1,ml);                    lcp=he[i+1];                    tp=min(lcp,ri-sa[i]+1);                    ml=max(ml,tp);                }            }            printf("%d\n",ans);        }    }    return 0;}

后缀自动机

#include<algorithm>#include<iostream>#include<string.h>#include<stdio.h>using namespace std;const int INF=0x3f3f3f3f;const int maxn=2010<<1;typedef long long ll;struct node{    node *f,*ch[30];    int len,way;    void init()    {        way=0;        memset(ch,0,sizeof ch);    }}*root,*tail,sam[maxn<<1];int tot,dp[2010][2010];char txt[maxn];void init(){    tot=0;    root=tail=&sam[tot++];    root->f=NULL;    root->len=0;    root->init();    root->way=1;}void Insert(int c){    node *p=tail,*np=&sam[tot++];    np->init(),np->len=p->len+1;    while(p&&!p->ch[c])    {        np->way+=p->way;        p->ch[c]=np,p=p->f;    }    tail=np;    if(!p)        np->f=root;    else    {        if(p->ch[c]->len==p->len+1)            np->f=p->ch[c];        else        {            node *q=p->ch[c],*nq=&sam[tot++];            *nq=*q;            nq->len=p->len+1;            nq->way=0;            q->f=np->f=nq;            while(p&&p->ch[c]==q)            {                p->ch[c]->way-=p->way;                nq->way+=p->way;                p->ch[c]=nq,p=p->f;            }        }    }}int main(){    int i,j,t,len,q,le,ri;    scanf("%d",&t);    while(t--)    {        scanf("%s",txt);        len=strlen(txt);        for(i=0; i<len; i++)        {            init();            for(j=i; j<len; j++)            {                Insert(txt[j]-'a');                dp[i][j]=tail->way;            }            for(j=i+1; j<len; j++)                dp[i][j]+=dp[i][j-1];        }        scanf("%d",&q);        while(q--)        {            scanf("%d%d",&le,&ri);            printf("%d\n",dp[le-1][ri-1]);        }    }    return 0;}

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