BZOJ3600：没有人的算术

传送门
如果能给每个 \(pair\) 按照权值编号就好了
假设之前已经有了所有的权值的编号，现在考虑编号新的 \(pair\)
如果看过了陈立杰的论文的话，不难得到一个重量平衡树的做法
给树上每个子树一个实数权值区间 \([l,r]\)，这个点权值为 \(mid=\frac{l+r}{2}\)
左子树 \([l,mid]\) 右子树 \([mid,r]\)
只需要选择一个树高 \(log\) 的树(treap/替罪羊树)使得满足精度要求即可

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(5e5 + 5);
const double alpha(0.75);

int ls[maxn], rs[maxn], rt, tot, size[maxn], que[maxn], cnt, id[maxn], n, m;
double val[maxn];
pair <int, int> info[maxn];
int mx[maxn << 2];

inline int operator <(pair <int, int> a, pair <int, int> b) {
	return val[a.first] == val[b.first] ? val[a.second] < val[b.second] : val[a.first] < val[b.first];
}

void Dfs(int u) {
	if (!u) return;
	Dfs(ls[u]), que[++cnt] = u, Dfs(rs[u]);
}

int Build(int l, int r, double vl, double vr) {
	if (l > r) return 0;
	double midv;
	int mid, o;
	mid = (l + r) >> 1, o = que[mid], midv = (vl + vr) * 0.5;
	ls[o] = rs[o] = 0, val[o] = midv;
	ls[o] = Build(l, mid - 1, vl, midv);
	rs[o] = Build(mid + 1, r, midv, vr);
	size[o] = size[ls[o]] + size[rs[o]] + 1;
	return o;
}

int Rebuild(int x, double vl, double vr) {
	cnt = 0, Dfs(x);
	return Build(1, cnt, vl, vr);
}

int Insert(int &x, double vl, double vr, pair <int, int> v) {
	double midv;
	int ret;
	midv = (vl + vr) * 0.5;
	if (!x) {
		x = ++tot, val[x] = midv, info[x] = v, size[x] = 1;
		return x;
	}
	if (alpha * size[x] < max(size[ls[x]], size[rs[x]])) x = Rebuild(x, vl, vr);
	if (v == info[x]) return x;
	else if (v < info[x]) ret = Insert(ls[x], vl, midv, v);
	else ret = Insert(rs[x], midv, vr, v);
	size[x] = size[ls[x]] + size[rs[x]] + 1;
	return ret;
}

void Modify(int x, int l, int r, int p) {
	int mid;
	if (l == r) mx[x] = l;
	else {
		mid = (l + r) >> 1;
		p <= mid ? Modify(x << 1, l, mid, p) : Modify(x << 1 | 1, mid + 1, r, p);
		mx[x] = val[id[mx[x << 1]]] >= val[id[mx[x << 1 | 1]]] ? mx[x << 1] : mx[x << 1 | 1];
	}
}

int Query(int x, int l, int r, int ql, int qr) {
	int mid, ret, v;
	if (ql <= l && qr >= r) return mx[x];
	mid = (l + r) >> 1, ret = -1, v;
	if (ql <= mid) ret = Query(x << 1, l, mid, ql, qr);
	if (qr > mid) {
		v = Query(x << 1 | 1, mid + 1, r, ql, qr);
		ret = (ret == -1 || val[id[v]] > val[id[ret]]) ? v : ret;
	}
	return ret;
}

int main() {
	int i, l, r, k;
	char op;
	scanf("%d%d", &n, &m);
	val[0] = -1, Insert(rt, 0, 1, make_pair(0, 0));
	for (i = 1; i <= n; ++i) Modify(1, 1, n, i);
	for (i = 1; i <= m; ++i) {
		scanf(" %c%d%d", &op, &l, &r);
		if (op == 'C') {
			scanf("%d", &k);
			id[k] = Insert(rt, 0, 1, make_pair(id[l], id[r]));
			Modify(1, 1, n, k);
		}
		else printf("%d\n", Query(1, 1, n, l, r));
	}
    return 0;
}