luogu解题报告：HNOI2012永无乡

https://www.luogu.org/problem/show?pid=3224

启发式合并+平衡树+并查集…
复杂度O(nlg2n)，还是可以过的..
Splay一次写过然而后面各种跪..调试能力捉急…估计是WC2017课前助眠音乐听太多了…

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

const int N = 100005;

struct node {
    int dat, siz, id;
    int chl[2], fa;
} tree[N*30];
int top = 0;

class splay_tree {
    int root;
    int new_node(int dat, int fa) 
    { //if (dat < 0) printf("!!%d-%d!!
", dat, tree[fa].id);
      return tree[++top].dat = dat, tree[top].fa = fa, tree[top].siz = 1, tree[top].chl[0] = tree[top].chl[1] = 0, top; }

    void update(int nd)
    { if (nd) tree[nd].siz = tree[tree[nd].chl[0]].siz + tree[tree[nd].chl[1]].siz + 1; }

    void zig(int nd) 
    {
        if (!nd || nd == root) return;
        int p = tree[nd].fa, tp = tree[p].chl[0] != nd, son = tree[nd].chl[tp^1];
        int g = tree[p].fa, tg = tree[g].chl[0] != p;
        tree[son].fa = p, tree[p].chl[tp] = son;
        tree[nd].chl[tp^1] = p, tree[p].fa = nd;
        tree[nd].fa = g;
        if (g) tree[g].chl[tg] = nd;
        else root = nd;
        update(p), update(nd);
    }

    void splay(int nd)
    {
        while (nd != root && tree[tree[nd].fa].fa) {
            int p = tree[nd].fa, g = tree[p].fa;
            int tp = tree[p].chl[0] != nd, tg = tree[g].chl[0] != p;
            if (tp == tg) zig(p), zig(nd);
            else zig(nd), zig(nd);
        }
        if (tree[nd].fa) zig(nd);
    }

    void insert(int &nd, int dat, int id, int fa = 0)
    {
        if (nd == 0) nd = new_node(dat, fa), tree[nd].id = id, splay(nd);
        else {
            //tree[nd].siz++;
            if (tree[nd].dat <= dat) insert(tree[nd].chl[1], dat, id, nd);
            else insert(tree[nd].chl[0], dat, id, nd);
            update(nd);
        }
    }

    void dfs(int nd, int tab = 0)
    {
        if (!nd) return;
        for (int i = 1; i <= tab; i++) putchar(' ');
        printf("id = %d, dat = %d, siz = %d
", tree[nd].id, tree[nd].dat, tree[nd].siz);
        dfs(tree[nd].chl[0], tab+2);
        dfs(tree[nd].chl[1], tab+2);
    }

public:

    splay_tree():root(0) {}
    inline void push(int dat, int id) 
    { insert(root, dat, id); }

    inline int& get_root() 
    { return root; }
    int find_kth(int k)
    { 
        //cout << tree[root].id << " " << k << " " << tree[root].siz << endl;
        if (k <= 0 || k > tree[root].siz) return 0;
        k--;
        int nd = root;
        while (nd > 0 && k >= 0) {
            //cout << tree[nd].id << " " << k << endl;
            //cout << tree[1026].id << endl;
            if (tree[tree[nd].chl[0]].siz == k) return nd;
            if (tree[tree[nd].chl[0]].siz > k) nd = tree[nd].chl[0];
            else k -= tree[tree[nd].chl[0]].siz + 1, nd = tree[nd].chl[1];
        }
        return 0;
    }

    void merge(int nd)
    {
        if (nd <= 0) return;
        //cout << nd << " " << tree[nd].dat << endl;
        push(tree[nd].dat, tree[nd].id);
        merge(tree[nd].chl[0]);
        merge(tree[nd].chl[1]);
    }

    void dfs()
    { dfs(root); }
} f[N];

int target[N];
inline int findp(int i) { return target[i]?target[i] = findp(target[i]):i;}
inline void link(int i, int j)
{ if (findp(i) != findp(j)) target[findp(i)] = findp(j); }

void merge(int a, int b)
{
    //cout << tree[f[a].get_root()].id << " f**k " << b << " " << findp(b) << " " << f[b].get_root() << " " << tree[f[b].get_root()].id << endl;
    if (f[a].get_root() <= 0 || f[b].get_root() <= 0) return;
    if (a != b) {
        if (tree[f[a].get_root()].siz < tree[f[b].get_root()].siz) f[b].merge(f[a].get_root()), f[a].get_root() = -1, link(a, b);
        else f[a].merge(f[b].get_root()), f[b].get_root() = -1, link(b, a);
    }
}

int n, m, q, dat;
int a, b;
char opt;

void work()
{
    freopen("bzoj_2733.in", "r", stdin);
    freopen("bzoj_2733.out", "w", stdout);
    memset(target, 0, sizeof target);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &dat), f[i].push(dat, i);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &a, &b);
        merge(findp(a), findp(b));
    }
    scanf("%d", &q);
    for (int i = 1; i <= q; i++) {
        scanf("
%c %d %d", &opt, &a, &b);
        if (opt == 'Q') dat = f[findp(a)].find_kth(b), printf("%d
", dat==0?-1:tree[dat].id);
        else if (opt == 'B') merge(findp(a), findp(b));
        else if (opt == 'D') f[findp(a)].dfs();
        else if (opt == 'L') cout << (findp(a) == findp(b)) << endl;
    }
}

int main()
{
    work();
    return 0;
}