[BZOJ2733][HNOI2012]永无乡（平衡树+启发式合并）

首先，构建出n棵平衡树，每棵平衡树只有一个节点，第i棵平衡树只包含第i座岛的相关信息。然后使用并查集维护岛之间的连通关系，对于加边操作，如果并查集中点x和y不连通，那么就在并查集中连接点x和y，并把x和y所在的平衡树合并。否则不做任何操作。
而对于合并两棵平衡树，可以使用启发式合并，即把点数较少的平衡树中的点暴力合并到点数较多的平衡树上。实际上，这样操作，每个点被合并的次数不超过O(logn)。
证明：由于启发式合并是把点数较少的平衡树合并到点数较多的平衡树上，所以一个点每被合并一次，所在的平衡树点数就至少扩大一倍，即被合并的次数不超过O(logn)。
对于询问操作，就是平衡树的「求排名为k的数」的问题，就不多说了。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}inline char get() {    char c; while ((c = getchar()) != 'B' && c != 'Q');    return c;}const int N = 1e5 + 5;int n, m, q, f[N], fa[N], lc[N], rc[N], val[N], sze[N];int cx(int x) {    if (f[x] != x) f[x] = cx(f[x]);    return f[x];}bool zm(int x, int y) {    int ix = cx(x), iy = cx(y);    if (ix != iy) {f[iy] = ix; return 1;}    return 0;}int which(int x) {return rc[fa[x]] == x;}void upt(int x) {    sze[x] = 1;    if (lc[x]) sze[x] += sze[lc[x]];    if (rc[x]) sze[x] += sze[rc[x]];}void rotate(int x) {    int y = fa[x], z = fa[y], b = lc[y] == x ? rc[x] : lc[x];    if (z) (lc[z] == y ? lc[z] : rc[z]) = x;    fa[x] = z; fa[y] = x; if (b) fa[b] = y;    if (lc[y] == x) rc[x] = y, lc[y] = b;    else lc[x] = y, rc[y] = b; upt(y); upt(x);}void splay(int x) {    while (fa[x]) {        if (fa[fa[x]]) {            if (which(x) == which(fa[x])) rotate(fa[x]);            else rotate(x);        }        rotate(x);    }    upt(x);}void ins(int rt, int z) {    int x = rt, y;    while (x) {        y = x; sze[x]++;        if (val[z] < val[x]) x = lc[x];        else x = rc[x];    }    if (val[z] < val[y]) lc[fa[z] = y] = z;    else rc[fa[z] = y] = z;}void dfs(int y, int x) {    if (lc[x]) dfs(y, lc[x]);    if (rc[x]) dfs(y, rc[x]);    fa[x] = lc[x] = rc[x] = 0;    sze[x] = 1; ins(y, x);}void merg(int x, int y) {    if (!zm(x, y)) return;    splay(x); splay(y);    if (sze[x] < sze[y]) swap(x, y);    dfs(x, y);}int query(int x, int y) {    splay(x);    while (x) {        int tmp = lc[x] ? sze[lc[x]] : 0;        if (y == tmp + 1) return x;        else if (y <= tmp) x = lc[x];        else x = rc[x], y -= tmp + 1;    }    return -1;}int main() {    int i, x, y; n = read(); m = read(); char op;    for (i = 1; i <= n; i++) f[i] = i, sze[i] = 1, val[i] = read();    while (m--) x = read(), y = read(), merg(x, y); q = read();    while (q--) {        op = get(); x = read(); y = read();        if (op == 'B') merg(x, y);        else printf("%d\n", query(x, y));    }    return 0;}