BZOJ2733 [HNOI2012]永无乡

直接平衡树启发式合并就好了。。。貌似是个很高端的东西。。

貌似可以证明splay的启发式合并是均摊$O(nlogn)$的。。。而其他平衡树都不行，所以其他的复杂度都是$O(nlog^2n)的$的

所以就用平板电视里的splay好啦！2333

 /**************************************************************
     Problem: 2733
     User: rausen
     Language: C++
     Result: Accepted
     Time:2148 ms
     Memory:9756 kb
 ****************************************************************/

 #include <cstdio>
 #include <ext/pb_ds/assoc_container.hpp>
 #include <ext/pb_ds/tree_policy.hpp>

 using namespace std;
 using namespace __gnu_pbds;
 typedef tree<int, int, less<int>, splay_tree_tag, tree_order_statistics_node_update> Tree;
 typedef Tree :: iterator iter;
 const int N = 1e5 + ;

 int read();
 int get_op();

 int n, m;
 int a[N];
 int fa[N], sz[N];
 Tree T[N];

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

 void _union(int x, int y) {
     iter it;
     x = find(x), y = find(y);
     if (x == y) return;
     if (sz[x] < sz[y]) swap(x, y);
     for (it = T[y].begin(); it != T[y].end(); ++it)
         T[x][it -> first] = it -> second;
     fa[y] = x, sz[x] += sz[y], T[y].clear();
 }

 int query() {
     int x = find(read()), k = read();
     if (k > T[x].size()) return -;
     return T[find(x)].find_by_order(k - ) -> second;
 }

 int main() {
     int i, Q, oper;
     n = read(), m = read();
     for (i = ; i <= n; ++i) {
         a[i] = read(), fa[i] = i;
         T[i][a[i]] = i;
     }
     for (i = ; i <= m; ++i) _union(read(), read());
     for (Q = read(); Q; --Q) {
         oper = get_op();
         if (oper == ) _union(read(), read());
         else printf("%d\n", query());
     }
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }

 inline int get_op() {
     static char ch;
     ch = getchar();
     while (ch != 'Q' && ch != 'B') ch = getchar();
     return ch == 'Q';
 }