CF763E Timofey and our friends animals [回滚莫队,可撤销并查集]
题意:
给你 \(n\) 个点,已知 \(m\) 对关系 \([u,v] (|u - v| \leq k)\),\(k\) 给出,询问 \(q\) 次,每次问你 \([l,r]\) 有多少个连通块。
sol:
显然的回滚莫队,我们按块来分,对于每个块 \(i\),我们可以把所有 \(l\in block_i\) 丢到 \(i\) 里面,然后我们对 \(r\) 排序,右指针向右移动,然后左指针那边撤销就好了,复杂度 \(n \sqrt n \log n\),如果在同一块的话直接暴力撤销就好了。
// 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;
using ull = unsigned long long;
using pii = pair<int, int>;
#define fir first
#define sec second
template <class T>
void cmax(T& x, const T& y) {
if (x < y) x = y;
}
template <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
template <class T>
void sort(vector<T>& v) {
sort(all(v));
}
template <class T>
void reverse(vector<T>& v) {
reverse(all(v));
}
template <class T>
void unique(vector<T>& v) {
sort(all(v)), v.erase(unique(all(v)), v.end());
}
void reverse(string& s) { reverse(s.begin(), s.end()); }
const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
char ch;
#ifndef __WIN64
char getchar() {
static char buf[io_size], *p1 = buf, *p2 = buf;
return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2) ? EOF : *p1++;
}
#endif
io_in& operator>>(char& c) {
for (c = getchar(); isspace(c); c = getchar())
;
return *this;
}
io_in& operator>>(string& s) {
for (s.clear(); isspace(ch = getchar());)
;
if (!~ch) return *this;
for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
;
return *this;
}
io_in& operator>>(char* str) {
char* cur = str;
while (*cur) *cur++ = 0;
for (cur = str; isspace(ch = getchar());)
;
if (!~ch) return *this;
for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
;
return *++cur = 0, *this;
}
template <class T>
void read(T& x) {
bool f = 0;
while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);
x = ~ch ? (ch ^ 48) : 0;
while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
x = f ? -x : x;
}
io_in& operator>>(int& x) { return read(x), *this; }
io_in& operator>>(ll& x) { return read(x), *this; }
io_in& operator>>(uint& x) { return read(x), *this; }
io_in& operator>>(ull& x) { return read(x), *this; }
io_in& operator>>(db& x) {
read(x);
bool f = x < 0;
x = f ? -x : x;
if (ch ^ '.') return *this;
double d = 0.1;
while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
return x = f ? -x : x, *this;
}
} in;
struct io_out {
char buf[io_size], *s = buf;
int pw[233], st[233];
io_out() {
set(7);
rep(i, pw[0] = 1, 9) pw[i] = pw[i - 1] * 10;
}
~io_out() { flush(); }
void io_chk() {
if (s - buf > io_limit) flush();
}
void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }
io_out& operator<<(char c) { return *s++ = c, *this; }
io_out& operator<<(string str) {
for (char c : str) *s++ = c;
return io_chk(), *this;
}
io_out& operator<<(char* str) {
char* cur = str;
while (*cur) *s++ = *cur++;
return io_chk(), *this;
}
template <class T>
void write(T x) {
if (x < 0) *s++ = '-', x = -x;
do {
st[++st[0]] = x % 10, x /= 10;
} while (x);
while (st[0]) *s++ = st[st[0]--] ^ 48;
}
io_out& operator<<(int x) { return write(x), io_chk(), *this; }
io_out& operator<<(ll x) { return write(x), io_chk(), *this; }
io_out& operator<<(uint x) { return write(x), io_chk(), *this; }
io_out& operator<<(ull x) { return write(x), io_chk(), *this; }
int len, lft, rig;
void set(int _length) { len = _length; }
io_out& operator<<(db x) {
bool f = x < 0;
x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
}
} out;
#define int long long
template <int sz, int mod>
struct math_t {
math_t() {
fac.resize(sz + 1), ifac.resize(sz + 1);
rep(i, fac[0] = 1, sz) fac[i] = fac[i - 1] * i % mod;
ifac[sz] = inv(fac[sz]);
Rep(i, sz - 1, 0) ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
vector<int> fac, ifac;
int qpow(int x, int y) {
int ans = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ans = ans * x % mod;
return ans;
}
int inv(int x) { return qpow(x, mod - 2); }
int C(int n, int m) {
if (n < 0 || m < 0 || n < m) return 0;
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
};
int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int lcm(int x, int y) { return x * y / gcd(x, y); }
const int maxn = 1e5 + 51;
const int S = 500;
int n, k, _;
struct qry {
int l, r, id;
bool operator<(const qry& other) const { return r < other.r; }
};
vector<qry> v[S];
bool L[maxn][6], R[maxn][6];
struct dsu {
int fa[maxn], sz[maxn];
int st[maxn], top = 0;
void init() { rep(i, 1, maxn - 1) sz[fa[i] = i] = 1; }
int find(int x) { return fa[x] == x ? x : find(fa[x]); }
void clear() {
while (top) {
int x = st[top--];
sz[fa[x]] -= sz[x];
fa[x] = x;
}
}
void merge(int x, int y, bool ins = 1) {
if (sz[x] < sz[y]) x ^= y ^= x ^= y;
sz[x] += sz[y];
fa[y] = x;
if (ins) st[++top] = y;
}
} dsu;
int bel(int x) { return (x - 1) / S + 1; }
int ans[maxn];
void solve(const qry& now) {
const int l = now.l;
const int r = now.r;
const int id = now.id;
int cnt = r - l + 1;
rep(i, l, r) {
for (int j = 1; j <= k && i + j <= r; j++) {
if (R[i][j]) {
int x = dsu.find(i);
int y = dsu.find(i + j);
if (x ^ y) {
cnt--;
dsu.merge(x, y);
}
}
}
}
ans[id] = cnt;
dsu.clear();
}
signed main() {
// code begin.
in >> n >> k >> _;
dsu.init();
while (_--) {
int u, v;
in >> u >> v;
if (u > v) u ^= v ^= u ^= v;
L[v][v - u] = R[u][v - u] = 1;
}
in >> _;
rep(i, 1, _) {
qry now;
in >> now.l >> now.r;
now.id = i;
if (now.r - now.l <= S)
solve(now);
else
v[bel(now.l)].pb(now);
}
rep(i, 1, bel(n)) {
if (!sz(v[i])) continue;
sort(v[i]);
dsu.init();
int r = i * S, cnt = 0;
for (auto x : v[i]) {
const int ql = x.l;
const int qr = x.r;
const int id = x.id;
while (r < qr) {
++r;
for (int j = 1; j <= k && r - j >= i * S; j++) {
if (L[r][j]) {
int x = dsu.find(r);
int y = dsu.find(r - j);
if (x ^ y) {
++cnt;
dsu.merge(x, y, 0);
}
}
}
}
int now = cnt;
for (int l = i * S - 1; l >= ql; l--) {
rep(j, 1, k) {
if (R[l][j]) {
int x = dsu.find(l);
int y = dsu.find(l + j);
if (x ^ y) {
++cnt;
dsu.merge(x, y);
}
}
}
}
ans[id] = qr - ql + 1;
ans[id] -= cnt;
cnt = now;
dsu.clear();
}
}
rep(i, 1, _) out << ans[i] << '\n';
return 0;
// code end.
}