HDU

Reincarnation

 HDU - 4622

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(sl...rl...r), sl...rl...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          

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.

Source

HDU - 4622

My Solution

题意：给出一个长度<=2000的字符串，然后给出q<=10000个询问，为s[l,r]内有多少个不同的子串。

后缀自动机

首先n <= 2000，故可以用O(n^2)的算法，故对于每个 i ~ n-1建立n个后缀自动机，每次添加一个字符就多出val[np] - val[pre[np]]个不同的子串，

valx]表示x节点存储的字符串个数即能表示的最长子串。

故用sum[maxn][maxn] n个前缀和来存储区间内不同子串的个数。

复杂度 O(n^2)

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;typedef long long LL;const int maxn = 2*2e3 + 8;string s;struct SAM{    int ch[maxn][26], pre[maxn], val[maxn];    int last, tot;    void init(){        last = tot = 0;        memset(ch[0], -1, sizeof ch[0]);        pre[0] = -1; val[0] = 0;    }    int extend(int c){        int p = last, np = ++tot;        val[np] = val[p] + 1;        memset(ch[np], -1, sizeof ch[np]);        while(~p && ch[p][c] == -1) ch[p][c] = np, p = pre[p];        if(p == -1) pre[np] = 0;        else{            int q = ch[p][c];            if(val[q] != val[p] + 1){                int nq = ++tot;                memcpy(ch[nq], ch[q], sizeof ch[q]);                val[nq] = val[p] + 1;                pre[nq] = pre[q];                pre[q] = pre[np] = nq;                while(~p && ch[p][c] == q) ch[p][c] = nq, p = pre[p];            }            else pre[np] = q;        }        last = np;        return val[np] - val[pre[np]];    }} sam;int sum[maxn][maxn];int main(){    #ifdef LOCAL    freopen("13.in", "r", stdin);    //freopen("13.out", "w", stdout);    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int T, n, i, j, q;    cin >> T;    while(T--){        cin >> s;        n = s.size();        for(i = 0; i < n; i++){            sam.init(); sum[i+1][0] = 0;            for(j = i; j < n; j++) sum[i+1][j+1] = sum[i+1][j] + sam.extend(s[j] - 'a');        }        cin >> q;        while(q--){            cin >> i >> j;            cout << sum[i][j] << "\n";        }    }    return 0;}

Thank you!

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