Luogu P4094 [HEOI2016/TJOI2016]字符串

题目描述

https://www.luogu.com.cn/problem/P4094

简要题意:给定一个长度为 $n$ 的字符串和 $m$ 次询问,每次询问给定四个参数 $a,b,c,d$,求 $S[a\cdots b]$ 的所有子串和 $S[c\cdots d]$ 的 $lcp$

$n,m\le 10^5$

Solution

首先考虑二分答案 $x$,然后我们考虑后缀数组,相当于求是否存在 $[a,b-x+1]$ 之间的后缀,满足排序后这个后缀到 $c$ 的区间最小值大于等于 $x$

我们考虑先二分出 $c$ 周围的最小值大于 $x$ 的区间 $[l,r]$,那么就相当于二维区间查询,主席树显然可以胜任

时间复杂度 $O(n\log^2n)$

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#include <iostream>
#define maxn 100010
using namespace std;

int n, m;
char s[maxn];

int tax[maxn], tp[maxn], sa[maxn], rk[maxn], M = 255;
void rsort() {
for (int i = 0; i <= M; ++i) tax[i] = 0;
for (int i = 1; i <= n; ++i) ++tax[rk[i]];
for (int i = 1; i <= M; ++i) tax[i] += tax[i - 1];
for (int i = n; i; --i) sa[tax[rk[tp[i]]]--] = tp[i];
}

int H[maxn];
void init_sa() {
if (n == 1) return sa[1] = rk[1] = 1, void(); int cnt = 1;
for (int i = 1; i <= n; ++i) rk[i] = s[i], tp[i] = i; rsort();
for (int k = 1; k < n; k *= 2) {
if (cnt == n) break; M = cnt; cnt = 0;
for (int i = n - k + 1; i <= n; ++i) tp[++cnt] = i;
for (int i = 1; i <= n; ++i) if (sa[i] > k) tp[++cnt] = sa[i] - k;
rsort(); swap(rk, tp); rk[sa[1]] = cnt = 1;
for (int i = 2; i <= n; ++i) {
if (tp[sa[i - 1]] != tp[sa[i]] || tp[sa[i - 1] + k] != tp[sa[i] + k]) ++cnt;
rk[sa[i]] = cnt;
}
} int lcp = 0;
for (int i = 1; i <= n; ++i) {
if (lcp) --lcp;
int j = sa[rk[i] - 1];
while (s[j + lcp] == s[i + lcp]) ++lcp;
H[rk[i]] = lcp;
}
}

int Log[maxn], st[maxn][21];
void init_st() { Log[0] = -1;
for (int i = 1; i <= n; ++i) Log[i] = Log[i >> 1] + 1, st[i][0] = H[i];
for (int j = 1; j <= 20; ++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]);
}

int lcp(int l, int r) {
if (++l > r) return n - sa[r] + 1;
int k = Log[r - l + 1];
return min(st[l][k], st[r - (1 << k) + 1][k]);
}

#define lc T[i].ch[0]
#define rc T[i].ch[1]
#define Lc T[j].ch[0]
#define Rc T[j].ch[1]
struct zhuxi {
int v, ch[2];
} T[maxn * 20]; int rt[maxn], top;
void update(int &i, int j, int l, int r, int k, int v) {
i = ++top; T[i] = T[j]; T[i].v += v;
if (l == r) return ; int m = l + r >> 1;
if (k <= m) update(lc, Lc, l, m, k, v);
else update(rc, Rc, m + 1, r, k, v);
}

int query(int i, int j, int l, int r, int L, int R) {
if (l > R || r < L) return 0;
if (L <= l && r <= R) return T[j].v - T[i].v;
int m = l + r >> 1;
return query(lc, Lc, l, m, L, R) + query(rc, Rc, m + 1, r, L, R);
}

bool check(int x, int a, int b, int c) {
int l = 1, r = rk[c], mid, L, R;
while (l <= r) {
mid = l + r >> 1;
if (lcp(mid, rk[c]) >= x) L = mid, r = mid - 1;
else l = mid + 1;
}

l = rk[c]; r = n;
while (l <= r) {
mid = l + r >> 1;
if (lcp(rk[c], mid) >= x) R = mid, l = mid + 1;
else r = mid - 1;
}

return query(rt[a - 1], rt[b - x + 1], 1, n, L, R) > 0;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m >> s + 1; init_sa(); init_st();
for (int i = 1; i <= n; ++i) update(rt[i], rt[i - 1], 1, n, rk[i], 1);
for (int i = 1; i <= m; ++i) {
int a, b, c, d; cin >> a >> b >> c >> d;
int l = 1, r = min(b - a + 1, d - c + 1), mid, ans = 0;
while (l <= r) {
mid = l + r >> 1;
if (check(mid, a, b, c)) ans = mid, l = mid + 1;
else r = mid - 1;
} cout << ans << "\n";
}
return 0;
}