[题解] bzoj 3600 没有人的算数 (替罪羊树+线段树)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cstdlib>
#define pii pair<int, int>
#define mp make_pair
#define ft first
#define sd second
#define ls C[rt][0]
#define rs C[rt][1]
using namespace std;

template <typename ty> void read(ty &x) {
  x = 0; int f = 1; char ch = getchar();
  while (ch > '9' || ch < '0') { if (ch == '-') f = -1; ch = getchar(); }
  while (ch >= '0' && ch <= '9') { x = x*10 + ch - '0'; ch = getchar(); }
  x *= f;
}
template <typename ty> ty Max(ty a, ty b) { return a > b ? a : b; }
template <typename ty> ty Min(ty a, ty b) { return a < b ? a : b; }
template <typename ty> int Chkmin(ty &a, ty b) { return a > b ? a = b, 1 : 0; }
template <typename ty> int Chkmax(ty &a, ty b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;
typedef double db;

const int inf = 0x7fffffff;
const int M = 5e5 + 16;
const int N = 1e5 + 16;
const db alpha = 0.76;
const LL MAXN = (1LL << 62) - 1;

//scapegoat_tree
LL V[M], L[M], R[M];
int C[M][2], F[M], S[M], A[M], B[M];
// A,B表示点对的前后两个点在替罪羊树中的编号
// 注意给0这个(伪)点对的A,B初始化为<1即可,因为点对都是由两个书中的点(序号大于0)构成的

//segment_tree
int D[N], T[N << 2];
// D表示数列中的点对应的点对在替罪羊树中的编号
// T表示线段树中区间的答案(指向一个原数列中的数)
int CUR[M];
int root, sz, n, m, l, r, k, tot;
int ans, num, mpt;
char CH[10];

bool blc(int rt) {
  return (db) S[rt] * alpha >= (db) S[ls]
    && (db) S[rt] * alpha >= (db) S[rs];
}

void init() {
  root = sz = 1;
  V[sz] = (1 + MAXN) >> 1;
  L[sz] = 1, R[sz] = MAXN;
  A[sz] = 0, B[sz] = 0;
}

void recycle(int rt) {
  if (!rt) return;
  recycle(ls);
  CUR[++tot] = rt;
  recycle(rs);
}

int build(int l, int r, LL x, LL y) {
  if (l > r) return 0;
  int mid = l + r >> 1, now = CUR[mid];
  L[now] = x, R[now] = y, V[now] = (x + y) >> 1;
  F[C[now][0] = build(l, mid - 1, x, V[now])] = now;
  F[C[now][1] = build(mid + 1, r, V[now], y)] = now;
  S[now] = S[C[now][0]] + S[C[now][1]] + 1;
  return now;
}

void rebuild(int rt) {
  LL l = L[rt], r = R[rt];
  int fa = F[rt], a = (C[fa][1] == rt);
  tot = 0; recycle(rt);
  int tmp = build(1, tot, l, r);
  if (root == rt) root = tmp;
  F[C[fa][a] = tmp] = fa;
}

int ins(int &rt, int pre, int x, int y) {
  if (!rt) {
    rt = ++ sz; A[sz] = x, B[sz] = y; S[sz] = 1;
    if (C[pre][0] == rt) L[sz] = L[pre], R[sz] = V[pre];
    else L[sz] = V[pre], R[sz] = R[pre];
    V[sz] = L[sz] + R[sz] >> 1; F[rt] = pre;
    return sz;
  }
  if (x == A[rt] && y == B[rt]) return rt;
  int tmp;
  if (V[x] > V[A[rt]] || (x == A[rt] && V[y] > V[B[rt]]))
    tmp = ins(C[rt][1], rt, x, y);
  else
    tmp = ins(C[rt][0], rt, x, y);
  S[rt] = S[ls] + S[rs] + 1;
  if (!blc(rt)) mpt = rt;
  return tmp;
}

int insert(int x, int y) {
  mpt = 0;
  int ans = ins(root, 0, x, y);
  if (mpt) rebuild(mpt);
  return ans;
}

void build(int rt, int l, int r) {
  if (l == r) { T[rt] = l; return; }
  int mid = l + r >> 1;
  build(rt << 1, l, mid); build(rt << 1 | 1, mid + 1, r);
  T[rt] = T[rt << 1];
}

int get_max(int a, int b) {
  int aa = A[D[a]], ab = A[D[b]];
  if (V[aa] < V[ab]) return b;
  if (V[aa] == V[ab] && V[B[D[a]]] < V[B[D[b]]]) return b;
  return a;
}

void modify(int rt, int l, int r, int p) {
  if (l == r) return;
  int mid = l + r >> 1;
  if (p <= mid) modify(rt << 1, l, mid, p);
  else modify(rt << 1 | 1, mid + 1, r, p);
  T[rt] = get_max(T[rt << 1], T[rt << 1 | 1]);
}

int query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) return T[rt];
  int mid = l + r >> 1;
  int a = 0, b = 0;
  if (x <= mid) a = query(rt << 1, l, mid, x, y);
  if (y > mid) b = query(rt << 1 | 1, mid + 1, r, x, y);
  if (!a || !b) return a + b;
  else return get_max(a, b);
}

int main () {

  read(n); read(m);
  init();
  for (int i = 1; i <= n; ++ i) D[i] = 1;
  build(1, 1, n);
  while (m --) {
    scanf("%s", CH);
    read(l); read(r);
    if (CH[0] == 'C') {
      read(k);
      D[k] = insert(D[l], D[r]);
      modify(1, 1, n, k);
    }
    else
      printf("%d\n", query(1, 1, n, l, r));
  }

  return 0;

}