【Codechef】【Chef and Graph Queries】Lct 可持久化线段树
2015-01-26 10:45
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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。
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 , Q, Test, b , val, q, X , Y , 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; }
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