传送门
Luogu
解题思路
很容易想到平衡树，然后还可以顺便维护一下连通性，但是如何合并两棵平衡树？
我们采用一种类似于启发式合并的思想，将根节点siz较小的那颗平衡树暴力的合并到另一颗上去。
那么复杂度呢？
由于一个点所在的平衡树在经过这样一次合并之后，根节点的siz至少乘2，所以每一次合并的复杂度是 (O(nlog n)) 的，所以整个算法的复杂度也就维持在了 (O(nlog n)) 级别，这题就搞定了。
细节注意事项
- 咕咕咕
参考代码
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
s = 0; int f = 0; char c = getchar();
while (!isdigit(c)) f |= (c == '-'), c = getchar();
while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
s = f ? -s : s;
}
const int _ = 100010;
int n, m, q, rt[_];
inline int findd(int x) { return rt[x] == x ? x : rt[x] = findd(rt[x]); }
int tot, val[_], pri[_], ls[_], rs[_], siz[_];
inline int newnode(int v)
{ return siz[++tot] = 1, pri[tot] = rand(), val[tot] = v, tot; }
inline void pushup(int p) { siz[p] = siz[ls[p]] + siz[rs[p]] + 1; }
inline void split(int p, int v, int& x, int& y) {
if (!p) return (void) (x = y = 0);
if (val[p] <= v)
x = p, split(rs[p], v, rs[p], y);
else
y = p, split(ls[p], v, x, ls[p]);
pushup(p);
}
inline int merge(int x, int y) {
if (!x || !y) return x + y;
if (pri[x] < pri[y])
return rs[x] = merge(rs[x], y), pushup(x), x;
else
return ls[y] = merge(x, ls[y]), pushup(y), y;
}
inline int kth(int p, int k) {
if (siz[ls[p]] + 1 > k) return kth(ls[p], k);
if (siz[ls[p]] + 1 == k) return p;
if (siz[ls[p]] + 1 < k) return kth(rs[p], k - siz[ls[p]] - 1);
}
inline void dfs(int x, int& y) {
if (!x) return;
dfs(ls[x], y);
dfs(rs[x], y);
int a, b;
split(y, val[x], a, b);
y = merge(merge(a, x), b);
}
inline int unionn(int x, int y) {
if (siz[x] > siz[y]) swap(x, y);
return dfs(x, y), y;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
#endif
srand(time(0)), read(n), read(m);
for (rg int v, i = 1; i <= n; ++i)
read(v), rt[i] = newnode(v);
for (rg int x, y, i = 1; i <= m; ++i) {
read(x), x = findd(x);
read(y), y = findd(y);
if (x == y) continue;
rt[x] = rt[y] = unionn(rt[x], rt[y]);
}
read(q);
char s[5];
for (rg int x, y, i = 1; i <= q; ++i) {
scanf("%s", s), read(x), read(y);
if (s[0] == 'Q') {
x = findd(x);
if (siz[x] < y) puts("-1");
else printf("%d
", kth(x, y));
} else {
x = findd(x), y = findd(y);
if (x == y) continue;
rt[x] = rt[y] = unionn(rt[x], rt[y]);
}
}
return 0;
}
完结撒花 (qwq)