BZOJ 3545: [ONTAK2010]Peaks

支持离线
边和询问都按权值排序，然后扫描线对每一条边进行处理，将小于等于当前询问权值的边进行处理
加入一条边相当于合并两个连通块，用并查集维护根，线段树合并两个根处的线段树即可
查询就在根查询权值线段树的第 \(k\) 大
把递归版的线段树换成循环的trie的写法，跑得更慢了。。

#include <bits/stdc++.h>
#define lp tree[p].l
#define rp tree[p].r
#define rq tree[q].r
#define lq tree[q].l
#define mid ((l + r) >> 1)

char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read() {
    int x = 0, f = 1; char ch = getc();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getc(); }
    return x * f;
}

const int N = 1e5 + 7;
struct Edge {
    int u, v, c;
    void in() { u = read(), v = read(), c = read(); }
    bool operator < (const Edge &p) const { return c < p.c; }   
} edge[N * 5];
struct Que {
    int u, c, k, id;
    void in() { u = read(), c = read(), k = read(); }
    bool operator < (const Que &p) const { return c < p.c; }
} Q[N * 5];
struct Node {
    int l, r, sum;
} tree[N * 30];

int fa[N], sz[N], n, m, h[N], w[N], q, root[N], ans[N * 5], tol;
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }

void update(int &p, int l, int r, int pos) {
    if (!p) p = ++tol;
    tree[p].sum++;
    if (l == r) return;
    if (pos <= mid) update(lp, l, mid, pos);
    else update(rp, mid + 1, r, pos);
}
int merge(int p, int q) {
    if (!p || !q) return p | q;
    tree[p].sum += tree[q].sum;
    lp = merge(lp, lq);
    rp = merge(rp, rq);
    return p;
}
int query(int p, int l, int r, int k) {
    if (!p) return -1;
    if (l == r) return l;
    if (tree[lp].sum >= k) return query(lp, l, mid, k);
    return query(rp, mid + 1, r, k - tree[lp].sum);
}

void dsu_merge(int u, int v) {
    u = find(u), v = find(v);
    if (u == v) return;
    if (sz[u] < sz[v]) std::swap(u, v);
    fa[v] = u; sz[u] += sz[v];
    root[u] = merge(root[u], root[v]);
}

int main() {
    n = read(), m = read(), q = read();
    for (int i = 1; i <= n; i++)
        h[i] = w[i] = read();
    std::sort(w + 1, w + 1 + n);
    int all = std::unique(w + 1, w + 1 + n) - w - 1;
    for (int i = 1; i <= m; i++) edge[i].in();
    for (int i = 1; i <= q; i++) Q[i].in(), Q[i].id = i;
    std::sort(edge + 1, edge + 1 + m);
    std::sort(Q + 1, Q + 1 + q);
    for (int i = 1; i <= n; i++) {
        fa[i] = i; sz[i] = 1;
        update(root[i], 1, all, std::lower_bound(w + 1, w + 1 + all, h[i]) - w);
    }
    int cur = 1;
    for (int i = 1; i <= q; i++) {
        while (cur <= m && edge[cur].c <= Q[i].c) {
            dsu_merge(edge[cur].u, edge[cur].v);
            cur++;
        }
        int f = find(Q[i].u);
        if (sz[f] < Q[i].k) { ans[Q[i].id] = -1; continue; }
        int k = sz[f] - Q[i].k + 1;
        ans[Q[i].id] = w[query(root[f], 1, all, k)];
    }
    for (int i = 1; i <= q; i++)
        printf("%d\n", ans[i]);
    return 0;
}

BZOJ 3545: [ONTAK2010]Peaks

原文：https://www.cnblogs.com/Mrzdtz220/p/12326091.html