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$\text{开}\mathop{\text{卷}}\limits^{ju\check{a}n}\text{有益}$

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P4094 [HEOI2016/TJOI2016] 字符串

P4094 [HEOI2016/TJOI2016] 字符串

题面

题目链接

解法

答案具有二分性,考虑先二分答案$len$并判断。

判断时在$height$数组上二分出$cl \le rk[c] \le cr$两个端点,使得$lcp(cl, cr) \ge len$,然后判断是否有$i$使得$i \in [a, b - len + 1]$且$rk[i] \in [cl, cr]$,显然可以用主席树维护查询。

时间复杂度$O(n \log^2 n)$,常数并不小。

AC代码

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/**
* @file: P4094.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P4094
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
const int MAXN = 1e5 + 10;
const int LOGN = 19;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct SuffixArray {
int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN], he[MAXN];
void radix_sort(int m) {
static int buc[MAXN];
for (int i = 0; i <= m; ++i)
buc[i] = 0;
for (int i = 1; i <= n; ++i)
buc[rk[i]]++;
for (int i = 1; i <= m; ++i)
buc[i] += buc[i - 1];
for (int i = n; i >= 1; --i)
sa[buc[rk[tp[i]]]--] = tp[i];
}
void init(int n, char* s) {
this->n = n;
int m = 200;
for (int i = 1; i <= n; ++i)
rk[i] = s[i] + 1, tp[i] = i;
radix_sort(m);
for (int w = 1, p = 0; p < n; m = p, w <<= 1) {
p = 0;
for (int i = 1; i <= w; ++i)
tp[++p] = n - w + i;
for (int i = 1; i <= n; ++i)
if (sa[i] > w)
tp[++p] = sa[i] - w;
radix_sort(m);
copy(rk + 1, rk + n + 1, tp + 1);
rk[sa[1]] = p = 1;
for (int i = 2; i <= n; ++i) {
if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
rk[sa[i]] = p;
else
rk[sa[i]] = ++p;
}
}
for (int i = 1, k = 0; i <= n; ++i) {
if (k)
k--;
while (s[i + k] == s[sa[rk[i] - 1] + k])
k++;
ht[i] = he[rk[i]] = k;
}
}
} sa;
struct SegmentTree {
struct Node {
int sum, ls, rs;
} nd[MAXN * 25];
int tot, rt[MAXN];
int& operator[](int i) { return rt[i]; }
int newnode(int i) { return nd[++tot] = nd[i], tot; }
void update(int& i, int l, int r, int q, int v) {
i = newnode(i);
if (l == r)
return void(nd[i].sum += v);
int mid = (l + r) >> 1;
if (q <= mid)
update(nd[i].ls, l, mid, q, v);
else
update(nd[i].rs, mid + 1, r, q, v);
nd[i].sum = nd[nd[i].ls].sum + nd[nd[i].rs].sum;
}
int query(int u, int v, int l, int r, int ql, int qr) {
if (ql <= l && r <= qr)
return nd[v].sum - nd[u].sum;
int mid = (l + r) >> 1, ret = 0;
if (ql <= mid)
ret += query(nd[u].ls, nd[v].ls, l, mid, ql, qr);
if (qr > mid)
ret += query(nd[u].rs, nd[v].rs, mid + 1, r, ql, qr);
return ret;
}
} sgt;
int n, m;
int lg2[MAXN], st[MAXN][LOGN];
char str[MAXN];
int lcp(int l, int r) {
if (l == r)
return INF;
int k = lg2[r - l];
return min(st[l + 1][k], st[r - (1 << k) + 1][k]);
}
bool check(int a, int b, int c, int d, int len) {
int l, r, cl, cr;
l = 1, r = sa.rk[c], cl = r;
while (l <= r) {
int mid = (l + r) >> 1;
if (lcp(mid, sa.rk[c]) >= len)
r = mid - 1, cl = mid;
else
l = mid + 1;
}
l = sa.rk[c], r = n, cr = l;
while (l <= r) {
int mid = (l + r) >> 1;
if (lcp(sa.rk[c], mid) >= len)
l = mid + 1, cr = mid;
else
r = mid - 1;
}
return sgt.query(sgt[a - 1], sgt[b - len + 1], 1, n, cl, cr);
}
signed main() {
read(n), read(m);
scanf("%s", str + 1);
sa.init(n, str);
for (int i = 2; i <= n; ++i)
lg2[i] = (i & (i - 1)) ? lg2[i - 1] : lg2[i - 1] + 1;
for (int i = 1; i <= n; ++i)
st[i][0] = sa.he[i];
for (int j = 1; j < LOGN; ++j) {
for (int i = 1; i + (1 << j) - 1 <= n; ++i) {
st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
}
for (int i = 1; i <= n; ++i)
sgt.update(sgt[i] = sgt[i - 1], 1, n, sa.rk[i], 1);
while (m--) {
int a, b, c, d;
read(a), read(b), read(c), read(d);
int l = 1, r = min(b - a + 1, d - c + 1), ans = 0;
while (l <= r) {
int mid = (l + r) >> 1;
if (check(a, b, c, d, mid))
l = mid + 1, ans = mid;
else
r = mid - 1;
}
write(ans), putchar('\n');
}
return 0;
}