hdu 4622

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

 

题意:

给出一个字符串,求出在每个区间内的字串的个数....

 

思路:

两个做法,SAM 和 SA.

先说SAM. SAM的一个非常实用的性质,SAM是在线增量构造的.所以,我们就能够在线构造每个区间所对应的SAM.

合并左端点相同的询问,然后在线构造SAM,构造出SAM后,用O(n)的时间内统计出字串就ok了. 跟SPOJ那道题方法差不多.

 

然后是SA

参考的沐阳的博客....写得非常有道理.

构造出SA之后,我们通过遍历在局部的后缀(起点在[L,R]之间)来计算字串的贡献.....

关键是我们如何通过全局的SA来推算局部的SA.

la 与 lb 分别为一前一后的后缀在区间内的长度.

 

如果 lcp < la && lcp < lb

那么 la 就是下一个后缀.

 

如果 lcp >= la && lcp >= lb

如果 la < lb 就要换后缀.

 

如果 lcp >= la &&　lcp < lb

就更换后缀．

 

这样下来我们就能够得出全局的SA

然后在中间维护一个rmq就能够快速求出lcp

问题就解决了.

 

#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define rep(i,a,b) for(i = a; i < b; ++i)
#define rev(i,a,b) for(i = a; i >= b; --i)
using namespace std;
const int maxn = (int)2500, inf = 0x3f3f3f3f;
int sa[maxn],rank[maxn],height[maxn],t1[maxn],t2[maxn],c[maxn];
char str[maxn];
int n;
int cmp(int x[],int a,int b,int c){
    return x[a] == x[b] && x[a+c] == x[b+c];
}
void build(int m){
    int i,j,p = 0, *x = t1, *y = t2;
    rep(i,0,m) c[i] = 0;
    rep(i,0,n) c[x[i] = str[i]]++;
    rep(i,1,m) c[i] += c[i-1];
    rev(i,n-1,0) sa[--c[x[i]]] = i;
    for(j = 1; j < n && p < n; j <<= 1, m = p){
        p = 0;
        rep(i,n-j,n) y[p++] = i;
        rep(i,0,n) if(sa[i] >= j) y[p++] = sa[i] - j;
        rep(i,0,m) c[i] = 0;
        rep(i,0,n) c[x[y[i]]]++;
        rep(i,1,m) c[i] += c[i-1];
        rev(i,n-1,0) sa[--c[x[y[i]]]] = y[i];
        for(swap(x,y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j) ? p - 1 : p++;
    }
}
void calc_height(){
    int i,j,k = 0;
    rep(i,0,n) rank[sa[i]] = i;
    for(i = 0; i < n; height[rank[i++]] = k)
        for(k ? k-- : 0, j = sa[rank[i] - 1]; str[i+k] == str[j+k] && min(i,j) + k < n; k++);
}
int l,r,lcp,last,ret;
int getlen(int st,int ed){
    return ed - st + 1;
}
int calc(int now,int pas){
    int la = getlen(sa[now],r), lb = getlen(sa[pas],r);    
    return pas ? la - min(lcp,min(lb,la)) : la;
}
int getnext(int now,int pas){
    if(!pas) return now;
    int la = getlen(sa[now],r), lb = getlen(sa[pas],r);
    if(lcp < min(la,lb)) return now;
    if(lcp >= la && lcp >= lb && la > lb) return now;
    if(lcp < la && lcp >= lb) return now;
    return pas;
}
void solve(){
    int i;
    ret = last = 0; lcp = inf;
    rep(i,0,n){
        if(l <= sa[i] && sa[i] <= r){
            if(last) lcp = min(lcp,height[i]);
            ret += calc(i,last);
            int t = getnext(i,last);
            if(t == i) lcp = getlen(sa[i],r), last = i;
        }
        else if(last) lcp = min(lcp, height[i]);
    }
    printf("%d\n",ret);
}
int main()
{
    freopen("rec.in","r",stdin);
    freopen("rec.out","w",stdout);
    int T,q;
    scanf("%d\n",&T);
    while(T--){
        scanf("%s",str);
        n = strlen(str);
        str[n++] = 'a' - 1;
        build(255);
        calc_height();
        scanf("%d\n",&q);
        while(q--){
            scanf("%d %d\n",&l,&r);
            l--; r--;
            solve();
        }
    }
    return 0;
}