【Codechef】【Chef and Graph Queries】Lct 可持久化线段树

Problem code: GERALD07

一个无向图，q次询问，每次询问留下li到ri的边有几个联通块。n, m, q <= 200000.

先预处理出每个边能替代之前最早的的边bi使其还是一棵树，用Lct维护。用可持久化线段树查询。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <string>#define Rep(i, x, y) for (int i = x; i <= y; i ++)#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)#define Dwn(i, x, y) for (int i = x; i >= y; i --)#define u t[x]#define v t[y]#define lc u.ch[0]#define rc u.ch[1]#define Lc t[lc]#define Rc t[rc]#define tp u.par#define Tp t[tp]#define su seg[x]#define sv seg[y]#define Slc seg[su.lch]using namespace std;typedef long long LL;const int N = 200000 * 4;struct arr {int ch[2], par, p, mx, vl; bool rv;void Clr(int x, int y) { ch[1] = ch[0] = par = rv = 0, p = x, vl = mx = y; }} t[N*2];struct Segtree {int l, r, vl, lch, rch;void Clr(int lx, int rx) { l = lx, r = rx, vl = lch = rch = 0; }} seg[N*30];int n, m, sz, tz, T[N], Q, Test, b[N], val, q, X[N], Y[N], ans;void Upd(int x) {if (!x) return ;int z = min(Lc.mx, Rc.mx);if (z >= u.vl) u.mx = u.vl, u.p = x;else (Lc.mx < Rc.mx) ? (u.mx = Lc.mx, u.p = Lc.p) : (u.mx = Rc.mx, u.p = Rc.p);}void PD(int x) { if (x && u.rv) Lc.rv ^= 1, Rc.rv ^= 1, swap(lc, rc), u.rv = 0; }bool d(int x) { return Tp.ch[1] == x; }bool Sch(int x, int &y) { return (y = tp) && (v.ch[1] == x || v.ch[0] == x); }void Sc(int x, int y, bool f) { u.ch[f] = y, v.par = x; }void Rot(int x) {int y = tp, z, f = d(x);if (Sch(y, z)) Sc(z, x, d(y));else u.par = z;Sc(y, u.ch[!f], f), Sc(x, y, !f), Upd(y);}void Splay(int x) {int y, z;PD(x);while (Sch(x, y)) {PD(v.par), PD(y), PD(x);if (!Sch(y, z)) Rot(x);else (d(x) == d(y)) ? (Rot(y), Rot(x)) : (Rot(x), Rot(x));} Upd(x);}int Access(int x) { int y = 0; for (; x; x = tp) Splay(x), rc = y, Upd(y = x); return y; }void Make(int x) { t[Access(x)].rv ^= 1, Splay(x); }int Find(int x) {for (x = Access(x); PD(x), lc; x = lc) ; return x;}void Link(int x, int y) { Make(x), tp = y; }void Cut(int x, int y) { Make(x), Access(y), Splay(y), tp = v.ch[0] = 0, Upd(y); }void Tqry(int x, int y) {Make(x); int z = Access(y);val = t[z].mx, q = t[z].p;}int Build(int l, int r) {int x = ++ tz, mid = (l + r) / 2;su.Clr(l, r);if (l == r) return x;su.lch = Build(l, mid), su.rch = Build(mid + 1, r);return x;}int Add(int y, int z) {int x = ++ tz, mid = (sv.l + sv.r) / 2; su = sv, su.vl ++;if (su.l == su.r) return x;if (z <= mid) su.lch = Add(sv.lch, z);else su.rch = Add(sv.rch, z);return x;}int Qry(int x, int z) {int l = su.l, r = su.r, mid = (l + r) / 2;if (l == r) return su.vl;if (z > mid) return Slc.vl + Qry(su.rch, z);else return Qry(su.lch, z);}int main(){scanf ("%d", &Test);while (Test --) {scanf ("%d%d%d", &n, &m, &Q);sz = n, tz = 0;memset(X, 0, sizeof(X));memset(Y, 0, sizeof(Y));Rep(i, 0, n) t[i].Clr(i, 1 << 28);Rep(i, 1, m) {int x, y;scanf ("%d%d", &x, &y);if (x == y) { b[i] = i; continue ; }if (Find(x) == Find(y)) {Tqry(x, y), b[i] = val;Cut(q, X[q]), Cut(q, Y[q]);} else b[i] = 0;t[++ sz].Clr(sz, 1 << 28);X[sz] = x, Y[sz] = y;Link(sz, x), Link(sz, y), t[sz].vl = i, Splay(sz);}T[0] = Build(0, m);Rep(i, 1, m) {T[i] = Add(T[i-1], b[i]);}Rep(i, 1, Q) {int x, y;scanf ("%d%d", &x, &y);ans = Qry(T[y], x-1);ans -= x - 1;printf ("%d\n", n - ans);}}    return 0;}