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;
}