『题解』Codeforces1142B Lynyrd Skynyrd

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Portal

Portal1: Codeforces

Portal2: Luogu

Description

Recently Lynyrd and Skynyrd went to a shop where Lynyrd bought a permutation (p) of length (n), and Skynyrd bought an array (a) of length (m), consisting of integers from (1) to (n).

Lynyrd and Skynyrd became bored, so they asked you (q) queries, each of which has the following form: "does the subsegment of (a) from the (l)-th to the (r)-th positions, inclusive, have a subsequence that is a cyclic shift of (p)?" Please answer the queries.

A permutation of length (n) is a sequence of (n) integers such that each integer from (1) to (n) appears exactly once in it.

A cyclic shift of a permutation ((p_1, p_2, ldots, p_n)) is a permutation ((p_i, p_{i + 1}, ldots, p_{n}, p_1, p_2, ldots, p_{i - 1})) for some (i) from (1) to (n). For example, a permutation ((2, 1, 3)) has three distinct cyclic shifts: ((2, 1, 3)), ((1, 3, 2)), ((3, 2, 1)).

A subsequence of a subsegment of array (a) from the (l)-th to the (r)-th positions, inclusive, is a sequence (a_{i_1}, a_{i_2}, ldots, a_{i_k}) for some (i_1, i_2, ldots, i_k) such that (l leq i_1 < i_2 < ldots < i_k leq r).

Input

The first line contains three integers (n), (m), (q) ((1 le n, m, q le 2 cdot 10^5)) — the length of the permutation (p), the length of the array (a) and the number of queries.

The next line contains (n) integers from (1) to (n), where the (i)-th of them is the (i)-th element of the permutation. Each integer from (1) to (n) appears exactly once.

The next line contains (m) integers from (1) to (n), the (i)-th of them is the (i)-th element of the array (a).

The next (q) lines describe queries. The (i)-th of these lines contains two integers (l_i) and (r_i) ((1 le l_i le r_i le m)), meaning that the (i)-th query is about the subsegment of the array from the (l_i)-th to the (r_i)-th positions, inclusive.

Output

Print a single string of length (q), consisting of (0) and (1), the digit on the (i)-th positions should be (1), if the subsegment of array (a) from the (l_i)-th to the (r_i)-th positions, inclusive, contains a subsequence that is a cyclic shift of (p), and (0) otherwise.

Sample Input1

3 6 3
2 1 3
1 2 3 1 2 3
1 5
2 6
3 5

Sample Output1

110

Sample Input2

2 4 3
2 1
1 1 2 2
1 2
2 3
3 4

Sample Output2

010

Hint

In the first example the segment from the (1)-st to the (5)-th positions is (1, 2, 3, 1, 2). There is a subsequence (1, 3, 2) that is a cyclic shift of the permutation. The subsegment from the (2)-nd to the (6)-th positions also contains a subsequence (2, 1, 3) that is equal to the permutation. The subsegment from the (3)-rd to the (5)-th positions is (3, 1, 2), there is only one subsequence of length (3) ((3, 1, 2)), but it is not a cyclic shift of the permutation.

In the second example the possible cyclic shifts are (1, 2) and (2, 1). The subsegment from the (1)-st to the (2)-nd positions is (1, 1), its subsequences are not cyclic shifts of the permutation. The subsegment from the (2)-nd to the (3)-rd positions is (1, 2), it coincides with the permutation. The subsegment from the (3) to the (4) positions is (2, 2), its subsequences are not cyclic shifts of the permutation.

Solution

我们可以先预处理出(a_i)在(p)序列中的前一个数为(mathrm{last}_i)。如果它能构成一个合法的循环序列，就代表它能够向前位移(n - 1)次(mathrm{last})。所以我们可以用倍增来解决。我们取一个最大的合法循环序列的头表示为(mathrm{b}_i)，那么最后的条件就是：

[max ^ {r} _ {i = l}{mathrm{b}_i} ge l ]

满足就输出(1)，否则输出(0)。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

const int MAXN = 1000005, MAXM = 30;
int n, m, q, l, r, a[MAXN], b[MAXN], p[MAXN], last[MAXN], pos[MAXN], st[MAXN][MAXM];
inline int calc_step(int x) {
    int s = 0;
    for (int i = 25; i >= 0; i--)
        if (s + (1 << i) < n) {
            x = st[x][i];
            s += 1 << i;
        }
    return x;
}
inline int query(int l, int r) {
    int x = (int)log2(r - l + 1);
    return max(st[l][x], st[r - (1 << x) + 1][x]);//询问ST表
}
int main() {
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &p[i]);
        pos[p[i]] = i;
    }
    for (int i = 1; i <= m; i++) {
        scanf("%d", &a[i]);
        if (pos[a[i]] == 1) st[i][0] = last[p[n]]; else st[i][0] = last[p[pos[a[i]] - 1]];
        last[a[i]] = i;
    }
    for (int j = 1; j <= 25; j++)
        for (int i = 1; i <= m; i++)
            st[i][j] = st[st[i][j - 1]][j - 1];
    for (int i = 1; i <= m; i++)
        b[i] = calc_step(i);
    memset(st, 0, sizeof(st));
    for (int i = 1; i <= m; i++)
        st[i][0] = b[i];
    for (int j = 1; j <= (int)log2(m); j++)
        for (int i = 1; i <= m - (1 << j) + 1; i++)
            st[i][j] = max(st[i][j - 1], st[i + (1 << j - 1)][j - 1]);//ST表
    for (int i = 1; i <= q; i++) {
        scanf("%d%d", &l, &r);
        if (query(l, r) >= l) printf("1"); else printf("0");
    }
    return 0;
}