SPOJ7258 SUBLEX - Lexicographical Substring Search(后缀自动机)

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 smallest?


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 < 2^31).

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.

 

会做但是不会写qwq,

思路很简单:

把SAM的转移边形成的DAG图求出来,然后跟主席树查第$k$大一样,贪心枚举每一位,这个节点的某个儿子的大小大于$k$了,说明要找的串在这个儿子里,

否则就把$k$减去节点大小,继续找

一开始弄不清楚DAG图求出来的$size$和前缀树求出来的$size$有啥区别。

大概就是:

对于每个节点来说,DAG图求出来的$size$表示有多少以该节点代表的字符为起点的子串

前缀树求出来的$size$表示该节点所代表状态的$right$集合大小是多少,也就是该状态出现了多少次

这题居然卡dfs mmp

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN = 180001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
char s[MAXN];
int opt, K, N;
int fa[MAXN], len[MAXN], ch[MAXN][27], siz[MAXN], tot = 1, last = 1, root = 1;
void insert(int x) {
    int now = ++tot, pre = last; last = now; len[now] = len[pre] + 1;
    siz[now] = 1;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) fa[now] = root;
    else {
        int q = ch[pre][x];
        if(len[q] == len[pre] + 1) fa[now] = q;
        else {
            int nows = ++tot; len[nows] = len[pre] + 1;
            memcpy(ch[nows], ch[q], sizeof(ch[q]));
            fa[nows] = fa[q]; fa[q] = fa[now] = nows;
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nows;
        }
    }
}    
void Query(int K) {
    int now = root;
    while(K) {
        if(now != root) K--;
        if(K <= 0) break;
        for(int i = 0; i <= 25; i++) 
            if(ch[now][i]) {
                if(siz[ch[now][i]] >= K) {putchar(i + 'a'); now = ch[now][i]; break; }
                else K -= siz[ch[now][i]];        
            }
    }
    puts("");
}
void Topsort() {
    static int A[MAXN], a[MAXN]; 
    for(int i = 1; i <= tot; i++) A[len[i]]++;
    for(int i = 1; i <= N; i++)   A[i] += A[i - 1];
    for(int i = tot; i >= 1; i--) a[A[len[i]]--] = i;
    for(int i = 1; i <= tot; i++) siz[i] = 1;
    for(int i = tot; i ; i--)
        for(int j = 0; j <= 25; j++)
            siz[a[i]] += siz[ch[a[i]][j]];    
}
int main() {
    scanf("%s", s + 1);
    N = strlen(s + 1);
    for(int i = 1; i <= N; i++) insert(s[i] - 'a');
    Topsort();
    int T = read();
    while(T--) {
        int K = read();
        Query(K);         
    }
    return 0; 
}

 

posted @ 2018-06-28 15:27 自为风月马前卒 阅读(...) 评论(...) 编辑 收藏

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