#6041. 「雅礼集训 2017 Day7」事情的相似度 [set启发式合并+树状数组扫描线]

时间:2023-03-09 22:58:13
#6041. 「雅礼集训 2017 Day7」事情的相似度 [set启发式合并+树状数组扫描线]

SAM 两个前缀的最长后缀等价于两个点的 \(len_{lca}\) , 题目转化为求 \(l \leq x , y \leq r\) , \(max\{len_{lca(x,y)}\}\)

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {
if (x < y)
x = y;
}
Tp<class T> void cmin(T& x, const T& y) {
if (x > y)
x = y;
}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); }
Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN {
char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
FILEIN& operator>>(char& c) {
while (isspace(c = GETC()))
;
return *this;
}
FILEIN& operator>>(string& s) {
while (isspace(ch = GETC()))
;
s = ch;
while (!isspace(ch = GETC())) s += ch;
return *this;
}
Tp<class T> void read(T& x) {
bool sign = 0;
while ((ch = GETC()) < 48) sign ^= (ch == 45);
x = (ch ^ 48);
while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48);
x = sign ? -x : x;
}
FILEIN& operator>>(int& x) { return read(x), *this; }
FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {
const static int LIMIT = 1 << 22;
char quq[SZ], ST[233];
int sz, O;
~FILEOUT() { flush(); }
void flush() {
fwrite(quq, 1, O, stdout);
fflush(stdout);
O = 0;
}
FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
FILEOUT& operator<<(string str) {
if (O > LIMIT)
flush();
for (char c : str) quq[O++] = c;
return *this;
}
Tp<class T> void write(T x) {
if (O > LIMIT)
flush();
if (x < 0) {
quq[O++] = 45;
x = -x;
}
do {
ST[++sz] = x % 10 ^ 48;
x /= 10;
} while (x);
while (sz) quq[O++] = ST[sz--];
}
FILEOUT& operator<<(int x) { return write(x), *this; }
FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
// #define int long long
int n, m;
const int maxn = 2e5 + 52;
vector<int> g[maxn];
void add(int u, int v) { g[u].pb(v); }
struct BIT {
int c[maxn];
int low(int x) { return x & -x; }
void add(int x, int y) {
for (; x; x ^= low(x)) cmax(c[x], y);
}
int qry(int x) {
int ans = 0;
for (; x <= n; x += low(x)) cmax(ans, c[x]);
return ans;
}
} bit;
vector<pii> chg[maxn];
struct SAM {
int cnt = 1, last = 1, ch[maxn][2], fa[maxn], len[maxn], t[maxn], rt[maxn];
void ins(int c, int pos) {
int p = last, np = last = ++cnt;
len[np] = len[p] + 1;
t[pos] = np;
for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
if (!p)
fa[np] = 1;
else {
int q = ch[p][c];
if (len[q] == len[p] + 1)
fa[np] = q;
else {
int nq = ++cnt;
len[nq] = len[p] + 1;
ch[nq][0] = ch[q][0], ch[nq][1] = ch[q][1];
fa[nq] = fa[q], fa[q] = fa[np] = nq;
for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
}
}
}
set<int> s[maxn];
void merge(int a, int b, int len) {
for (auto it = s[b].begin(); it != s[b].end(); ++it) {
s[a].insert(*it);
auto pre = s[a].find(*it), nxt = pre;
++nxt;
if (pre != s[a].begin())
pre--, chg[*it].pb(*pre, len);
if (nxt != s[a].end())
chg[*nxt].pb(*it, len);
s[a].erase(*it);
}
for (int x : s[b]) s[a].insert(x);
}
void dfs(int u) {
for (int v : g[u]) {
dfs(v); // if(! len[u]) continue ;
if (s[rt[u]].size() > s[rt[v]].size())
merge(rt[u], rt[v], len[u]);
else
merge(rt[v], rt[u], len[u]), rt[u] = rt[v];
}
}
void build() {
for (int i = 1; i <= n; i++) s[t[i]].insert(i);
for (int i = 2; i <= cnt; i++) add(fa[i], i);
for (int i = 2; i <= cnt; i++) rt[i] = i;
dfs(1);
}
} sam;
vector<pii> qr[maxn];
int ans[maxn];
signed main() {
// code begin.
in >> n >> m;
string s;
in >> s;
int pos = 0;
for (char c : s) sam.ins(c ^ '0', ++pos);
sam.build();
rep(i, 1, m) {
int l, r;
in >> l >> r;
qr[r].pb(l, i);
}
rep(i, 1, n) {
for (auto x : chg[i]) bit.add(x.first, x.second);
for (auto x : qr[i]) ans[x.second] = bit.qry(x.first);
}
rep(i, 1, m) out << ans[i] << '\n';
return 0;
// code end.
}