【 bzoj 3545 】 [ONTAK2010]Peaks - Treap快速合并

　　听说这题有在线做法……？YY了一下似乎要把各种东西都给可持久化掉。。。好麻烦。。。
　　反正这题没强制在线
　　先把询问按他给的那个阀值排序，然后搞像Kurscal的过程，按边权从小到大枚举边，每次合并连接的两个点所能到的权值集合，这个可以用平衡树的启发式合并实现。于是询问就在对应的平衡树里查K大就可以了。
　　根据势能分析，n棵只有1个节点的splay启发式合并复杂度是 O(nlogn) 的。如果同样级别treap直接启发式合并的话复杂度是 O(nlog2n) 的。但是某十多年前的论文证明了两棵大小分别为 n,m(n≤m) 的treap的合并是 O(nlognm) 的。于是同样级别的treap合并也是 O(nlogn) 的了。
　　treap的合并就是代码里写的那样。
　　时间复杂度 O(mlogm+nlogn)
　　我还以为能跑多快结果才rank21

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i = a , _ = b ; i <= _ ; i ++)
#define per(i,a,b) for (int i = a , _ = b ; i >= _ ; i --)

#define gprintf(...) //fprintf(stderr , __VA_ARGS__)

inline int rd() {
    char c = getchar();
    while (!isdigit(c) && c != '-') c = getchar() ; int x = 0 , f = 1;
    if (c == '-') f = -1; else x = c - '0';
    while (isdigit(c = getchar())) x = x * 10 + c - '0';
    return x * f;
}

inline int rnd() {
    static int rand_seed = 42071823;
    rand_seed += (rand_seed << 1 | 1);
    return rand_seed;
}

const int maxn = 1000007;
const int maxm = 1000007;

typedef int arr[maxn];
typedef int que[maxm];

typedef pair<int , int> pii;

arr fa , rk , h , ans;

pii H[maxn];

int n , m , q;

int find(int u) { return u == fa[u] ? u : fa[u] = find(fa[u]); }

void init_UFS() {
    rep (i , 1 , n) fa[i] = i , rk[i] = 1;
}

struct edge {
    int u , v , w;
    edge(int u = 0 , int v = 0 , int w = 0): u(u) , v(v) , w(w) { }
}E[maxm];

struct info {
    int v , x , k , id;
    info(int v = 0 , int x = 0 , int k = 0 , int id = 0):v(v) , x(x) , k(k) , id(id) { }
}Q[maxm];

inline bool cmp_edge(const edge a , const edge b) {
    return a.w < b.w;
}

inline bool cmp_info_key(const info a , const info b) {
    return a.x < b.x;
}

struct node {
    node *l , *r;
    int key , pri;
    int sz;

    node (int key = 0): key(key) , pri(rnd()) , l(NULL) , r(NULL) , sz(1) { }

    inline void upd() {
        sz = 1;
        if (l) sz += l->sz;
        if (r) sz += r->sz;
    }
}mem_pool[maxn];

struct Droot {
    node *l , *r;
    Droot (node *a = NULL , node *b = NULL): l(a) , r(b) { }
};

int mem_top;

inline node *newnode(int key) {
    node *u = &mem_pool[mem_top ++];
    *u = node(key);
    return u;
}

inline int Size(node *u) {
    return u ? u->sz : 0;
}

node *join(node *u , node *v) {
    if (!u) return v;
    if (!v) return u;
    if (u->pri < v->pri) {
        u->r = join(u->r , v);
        u->upd();
        return u;
    } else {
        v->l = join(u , v->l);
        v->upd();
        return v;
    }
}

Droot split(node *u , int v) {
    if (!u) return Droot();
    Droot t;
    if (u->key < v) {
        t = split(u->r , v);
        u->r = t.l , t.l = u;
    } else {
        t = split(u->l , v);
        u->l = t.r , t.r = u;
    }
    u->upd();
    return t;
}

node *merge(node *u , node *v) {
    if (!u) return v;
    if (!v) return u;
    if (u->pri > v->pri) swap(u , v);
    Droot t = split(v , u->key);
    u->l = merge(u->l , t.l);
    u->r = merge(u->r , t.r);
    u->upd();
    return u;
}

struct Treap {
    node *rt;

    void init(int key) {
        rt = newnode(key);
    }

    inline void merge_to(Treap&v) {
        rt = merge(rt , v.rt);
    }

    inline int kth(int k) {
        if (k <= 0 || k > rt->sz) return -1;
        k = rt->sz - k + 1;
        for (node *u = rt;;) {
            int t = Size(u->l) + 1;
            if (k == t) return u->key;
            else if (k < t) u = u->l;
            else k -= t , u = u->r;
        }
    }
}path[maxn];

void input() {
    n = rd() , m = rd() , q = rd();
    rep (i , 1 , n) h[i] = rd();
    rep (i , 1 , n) H[i] = pii(h[i] , i);
    rep (i , 1 , m) {
        int u = rd() , v = rd() , w = rd();
        E[i] = edge(u , v , w);
    }
    rep (i , 1 , q) {
        int u = rd() , x = rd() , k = rd();
        Q[i] = info(u , x , k , i);
    }
    sort(H + 1 , H + n + 1);
    rep (i , 1 , n) h[i] = lower_bound(H + 1 , H + n + 1 , pii(h[i] , i)) - H;
    init_UFS();
}

inline void link(int u , int v) {
    u = find(u) , v = find(v);
    if (u == v) return;
    if (rk[u] <  rk[v]) swap(u , v);
    if (rk[u] == rk[v]) rk[u] ++;
    path[u].merge_to(path[v]);
    fa[v] = u;
}

void solve() {
    sort(E + 1 , E + m + 1 , cmp_edge);
    sort(Q + 1 , Q + q + 1 , cmp_info_key);
    rep (i , 1 , n) path[i].init(h[i]);
    for (int i = 1 , j = 0 ; i <= q && j <= m ; i ++) {
        for ( ; j + 1 <= m && E[j + 1].w <= Q[i].x ; j ++)
            link(E[j + 1].u , E[j + 1].v);
        int u = find(Q[i].v);
        ans[Q[i].id] = path[u].kth(Q[i].k);
    }
    rep (i , 1 , q) printf("%d\n" , ans[i] == -1 ? -1 : H[ans[i]].first);
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.txt" , "r" , stdin);
    #endif
    input();
    solve();
    return 0;
}