Lexicographical Substring Search

Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form:

If all distinct substrings of string S were sorted lexicographically, which one will be the K-th

After knowing the huge number of questions Kinan will ask, Daniel figured out that he can't do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan's questions.

Example:

S = "aaa" (without quotes)
substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:
"a", "aa", "aaa".

Input

In the first line there is Kinan's string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K

Output

Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.

Example

Input:
aaa
2
2
3

Output:
aa
aaa

Edited: Some input file contains garbage at the end. Do not process them.

My Solution

題意：給一個長度不大於90000的字串，每次詢問它的所有不同子串中，字典序第k小的，詢問不大於500個。

字尾陣列

本來是找了一個字尾自動機的題，看了題以後覺得用字尾陣列做比較方便，

結果交了一發，發現自己的程式碼竟然是目前為止vj上這個題史上最快的，害怕

開心 Y ( ^ _ ^ ) Y，

首先用字串s，倍增演算法跑出 sa和height，

然後把詢問根據v[j].q的大小排序

然後對於每個i，(1<=i<=n)，

cnt +=  (n - sa[i]) - height[i]; 表示截止當前字尾，所能表示的不同子串的第cnt小的子串，

所以如果 cnt >= v[ptr].q 則 v[ptr].ansi = sa[i], v[ptr].anslen = (n - sa[i]) - (cnt - v[ptr].q);然後ptr++，直到不滿足cnt >= v[ptr].q這個條件，

然後開始下一個字尾。

得到答案以後在把v陣列按照ind排序，然後輸出 s.substr(v[i].ansi, v[i].anslen)即可。

本來史上最快是60ms，筆者自己的程式竟然只用了10ms重新整理了記錄，雖然題目不大難，但能重新整理記錄還是很開心的 ☺☺。

複雜度 O(nlogn)

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 9e4 + 8; //記得有時候要開2倍
int sa[maxn], height[maxn];
int _rank[maxn], t1[maxn], t2[maxn], c[maxn];
string s;
inline void get_sa(const int &n, int m)
{
    int i, k, *x = t1, *y = t2, p, j;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) ++ c[x[i] = s[i]];
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[-- c[x[i]]] = i;
    for(k = 1; k <= n; k <<= 1){
        p = 0;
        for(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++) c[i] = 0;
        for(i = 0; i < n; i++) ++ c[x[y[i]]];
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y), p = 1, x[sa[0]] = 0;
        for(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 ++;
        if(p >= n) break;
        m = p;
    }
    k = 0;
    for(i = 0; i < n; i++) _rank[sa[i]] = i;
    for(i = 0; i < n; i++){
        if(k) --k; if(!_rank[i]) continue;
        j = sa[_rank[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        height[_rank[i]] = k;
    }
}

inline void print(const int &n)
{
    for(int i = 1; i <= n; i++){ //sa and height is 1~n based
        //cout << i << " : " << _rank[sa[i]] << " " << sa[i] << endl;
        for(int j = sa[i]; j < n; j++){ //the context is 0~n-1 based
            cout << s[j];
        }
        cout << endl;
    }
    cout << endl;
}

struct p{
    int q, ansi, anslen, ind;
} v[maxn];

inline bool cmpq(const p &a, const p &b)
{
    return a.q < b.q;
}

inline bool cmpind(const p &a, const p &b)
{
    return a.ind < b.ind;
}


int main()
{
    #ifdef LOCAL
    freopen("14.in", "r", stdin);
    //freopen("14.out", "w", stdout);

    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);
    LL n, q, i, ptr = 0, cnt = 0;
    cin >> s >> q;
    n = s.size();
    get_sa(n+1, 256);
    //print(n);
    for(i = 0; i < q; i++){
        cin >> v[i].q;
        v[i].ind = i;
    }
    sort(v, v + q, cmpq);
    for(i = 1; i <= n; i++){
        cnt +=  (n - sa[i]) - height[i];
        while(v[ptr].q <= cnt){
            if(ptr >= q) break;
            v[ptr].ansi = sa[i], v[ptr].anslen = (n - sa[i]) - (cnt - v[ptr].q);
            ptr++;

        }
    }
    sort(v, v + q, cmpind);
    for(i = 0; i < q; i++) cout << s.substr(v[i].ansi, v[i].anslen) << "\n";

    return 0;
}