Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P6640 [BJOI2020] 封印

P6640 [BJOI2020] 封印

$t$ 建 sam ,$s$ 在 sam 上匹配,$w _ i$ 表示当前位置结束的最大匹配长度。

最后答案为 $\max _ {i = l} ^ r \min(i - l + 1, w _ i)$ 。二分答案 $mid$ ,那么有 $i - l + 1 \ge mid$ ,即 $i \ge l + mid - 1$ ,查询是否存在 $w _ i \ge mid, i \ge l + mid - 1$ ,可以用 ST 表。

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 4e5 + 10, K = 20;
char S[N], T[N];
int n, q, cnt = 1, last = cnt, lg[N], st[N][K];
struct Node { int p, len, nxt[26]; } tr[N];
void insert (int c)
{
int p = last, np = ++cnt;
tr[np].len = tr[p].len + 1;
for (; p && !tr[p].nxt[c]; p = tr[p].p) tr[p].nxt[c] = np;
last = np;
if (!p) return void(tr[np].p = 1);
int q = tr[p].nxt[c];
if (tr[q].len == tr[p].len + 1) return void(tr[np].p = q);
int nq = ++cnt;
tr[nq] = tr[q], tr[nq].len = tr[p].len + 1;
tr[q].p = tr[np].p = nq;
for (; p && tr[p].nxt[c] == q; p = tr[p].p) tr[p].nxt[c] = nq;
}
void init ()
{
for (int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
for (int k = 1; 1 << k <= n; ++k)
for (int i = 1; i + (1 << k) - 1 <= n; ++i)
st[i][k] = max(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
}
int query (int l, int r)
{
if (l > r) return 0;
int k = lg[r - l + 1];
return max(st[l][k], st[r - (1 << k) + 1][k]);
}
int main ()
{
scanf("%s%s", S + 1, T + 1);
n = strlen(T + 1);
for (int i = 1; i <= n; ++i) insert(T[i] - 'a');
n = strlen(S + 1);
for (int i = 1, p = 1, len = 0; i <= n; ++i)
{
int c = S[i] - 'a';
while (p && !tr[p].nxt[c]) len = tr[p = tr[p].p].len;
if (!p) p = 1, len = 0;
else p = tr[p].nxt[c], ++len;
st[i][0] = len;
}
init();
read(q);
for (int a, b; q; --q)
{
read(a, b);
int l = 1, r = n, res;
while (l <= r)
{
int mid = l + r >> 1;
query(mid + a - 1, b) >= mid ? (res = mid, l = mid + 1) : r = mid - 1;
}
write(res), puts("");
}
return 0;
}