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!!\n", 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\n", 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("\n%c %d %d", &opt, &a, &b);        if (opt == 'Q') dat = f[findp(a)].find_kth(b), printf("%d\n", 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;}