HDU4622:Reincarnation(后缀数组，求区间内不同子串的个数)

时间：2023-03-08 20:20:28
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
2

bbaba

5

3 4

2 2

2 5

2 4

1 4

baaba

5

3 3

3 4

1 4

3 5

5 5
Sample Output
3

1

7

5

8

1

3

8

5

1



Hint


      I won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.
Author
WJMZBMR
Source
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题意：
求区间内不同子串的个数

思路：
论文里面有求整个串的不同子串的个数。我们能够引申到这道题
对于整个串，我们的求法是全部子串数减去全部height的值，而height就是lcp
那么对于某个区间，我们仅仅要求出全部包括在这个区间的后缀。然后减去互相之间的lcp就可以
关键是我们要保持这个区间的后缀的字典序

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <stack>

#include <queue>

#include <map>

#include <set>

#include <vector>

#include <math.h>

#include <bitset>

#include <algorithm>

#include <climits>

using namespace std;

#define LS 2*i

#define RS 2*i+1

#define UP(i,x,y) for(i=x;i<=y;i++)

#define DOWN(i,x,y) for(i=x;i>=y;i--)

#define MEM(a,x) memset(a,x,sizeof(a))

#define W(a) while(a)

#define gcd(a,b) __gcd(a,b)

#define LL long long

#define N 2005

#define MOD 1000000007

#define INF 0x3f3f3f3f

#define EXP 1e-8

#define rank rank1

int wa[N],wb[N],wsf[N],wv[N],sa[N];

int rank[N],height[N],s[N],a[N];

char str[N],str1[N],str2[N];

#define F(x) ((x)/3+((x)%3==1?

0:tb))

#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)

//sa:字典序中排第i位的起始位置在str中第sa[i]

//rank:就是str第i个位置的后缀是在字典序排第几

//height：字典序排i和i-1的后缀的最长公共前缀

int cmp(int *r,int a,int b,int k)

{

    return r[a]==r[b]&&r[a+k]==r[b+k];

}

void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0

{

    int i,j,p,*x=wa,*y=wb,*t;

    for(i=0; i<m; i++)  wsf[i]=0;

    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;

    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];

    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;

    p=1;

    j=1;

    for(; p<n; j*=2,m=p)

    {

        for(p=0,i=n-j; i<n; i++)  y[p++]=i;

        for(i=0; i<n; i++)  if(sa[i]>=j)  y[p++]=sa[i]-j;

        for(i=0; i<n; i++)  wv[i]=x[y[i]];

        for(i=0; i<m; i++)  wsf[i]=0;

        for(i=0; i<n; i++)  wsf[wv[i]]++;

        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];

        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];

        t=x;

        x=y;

        y=t;

        x[sa[0]]=0;

        for(p=1,i=1; i<n; i++)

            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;

    }

}

void getheight(int *r,int n)//n不保存最后的0

{

    int i,j,k=0;

    for(i=1; i<=n; i++)  rank[sa[i]]=i;

    for(i=0; i<n; i++)

    {

        if(k)

            k--;

        else

            k=0;

        j=sa[rank[i]-1];

        while(r[i+k]==r[j+k])

            k++;

        height[rank[i]]=k;

    }

}

int Log[N];

int best[30][N];

void setLog()

{

    Log[0] = -1;

    for(int i=1; i<N; i++)

    {

        Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1 ;

    }

}

void RMQ(int n)  //初始化RMQ

{

    for(int i = 1; i <= n ; i ++) best[0][i] = height[i];

    for(int i = 1; i <= Log[n] ; i ++)

    {

        int limit = n - (1<<i) + 1;

        for(int j = 1; j <= limit ; j ++)

        {

            best[i][j] = min(best[i-1][j] , best[i-1][j+(1<<i>>1)]);

        }

    }

}

int LCP(int a,int b)  //询问a,b后缀的最长公共前缀

{

    a ++;

    int t = Log[b - a + 1];

    return min(best[t][a] , best[t][b - (1<<t) + 1]);

}

int t,n,m;

int solve(int l,int r,int n)

{

    int ans = (r-l+1)*(r-l+2)/2;

    int last = -1;

    int cnt = r-l+1;

    for(int i = 1; i<=n; i++)

    {

        if(!cnt) break;

        if(sa[i]<l || sa[i]>r) continue;

        cnt--;

        if(last == -1)

        {

            last = i;

            continue;

        }

        int a = last;

        int b = i;

        if(a>b) swap(a,b);

        int lcp = LCP(a,b);

        int la = r-sa[last]+1;//区间内该串的尾部

        int lb = r-sa[i]+1;

        if(la>=lb && lcp>=lb);//la包括lb了，那么就用la继续往后比較，借此保持字典序，来模拟得到该区间的全部height

        else last = i;

        ans-=min(lcp,min(la,lb));

    }

    return ans;

}

int main()

{

    int i,j,k,len,l,r;

    scanf("%d",&t);

    setLog();

    W(t--)

    {

        scanf("%s",str);

        scanf("%d",&m);

        len = strlen(str);

        for(i = 0; i<len; i++)

            s[i] = str[i]-'a'+1;

        s[len] = 0;

        getsa(s,sa,len+1,30);

        getheight(s,len);

        RMQ(len);

        while(m--)

        {

            scanf("%d%d",&l,&r);

            printf("%d\n",solve(l-1,r-1,len));

        }

    }

    return 0;

}

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