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这题是 \(LCT\) 维护子树信息中的 \(LCT\) 维护重心

Description

link

题意概述：给定一个森林，要求支持以下操作

1.链接两个点

2.求一个点所在树的重心

3.求所有重心编号的异或和

Solution

\[Begin\]

看到有链接和询问操作的题目，我们想到了\(LCT\)

首先是一些重心的性质，本题可以用到：

\(1.\) 点到树上所有点的距离和最小的那个点就是中心

\(2.\) 重心在添加一条边之后只会移动最多一条边的距离

\(3.\) 如果我们联通森林里的两棵树，那么新树的重心就在原两树重心的路径上

应该都由重心的定义理解啥的易证吧\(2333\)

然后我们在处理 \(2\) 操作的时候搞个并查集（\(findroot\)好像很慢）

处理 \(3\) 操作的时候直接在链上进行类似二分查找的东西，看两侧子树的大小关系

\[Q.A.D\]

\(P.s.\)博主知道应该是\(QED\)

Code

#include <bits/stdc++.h>
using namespace std;
#define int long long
namespace yspm {
inline int read() {
    int res = 0, f = 1;
    char k;
    while (!isdigit(k = getchar()))
        if (k == '-')
            f = -1;
    while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
    return res * f;
}
const int N = 3e5 + 10, inf = 1e15 + 10;
int f[N], c[N][2], s[N], st[N], si[N], n, m, fa[N];
bool r[N];
inline void push_up(int x) { return s[x] = s[c[x][1]] + s[c[x][0]] + si[x] + 1, void(); }
inline bool notroot(int x) { return c[f[x]][0] == x || c[f[x]][1] == x; }
inline void push_down(int x) {
    if (r[x]) {
        swap(c[x][0], c[x][1]);
        r[c[x][0]] ^= 1;
        r[c[x][1]] ^= 1;
    }
    return r[x] = 0, void();
}
inline void push_all(int x) {
    if (notroot(x))
        push_all(f[x]);
    push_down(x);
    return;
}
inline void rotate(int x) {
    int y = f[x], z = f[y], k = (c[y][1] == x), w = c[x][!k];
    if (notroot(y))
        c[z][c[z][1] == y] = x;
    c[x][!k] = y;
    c[y][k] = w;
    if (w)
        f[w] = y;
    f[y] = x;
    f[x] = z;
    return push_up(y);
}
inline void splay(int x) {
    push_all(x);
    while (notroot(x)) {
        int y = f[x], z = f[y];
        if (notroot(y))
            rotate((c[y][0] == x) ^ (c[z][0] == y) ? x : y);
        rotate(x);
    }
    return push_up(x);
}
inline void access(int x) {
    for (int y = 0; x; x = f[y = x]) {
        splay(x);
        si[x] += s[c[x][1]], si[x] -= s[c[x][1] = y];
        push_up(x);
    }
    return;
}
inline void makeroot(int x) {
    access(x);
    splay(x);
    r[x] ^= 1;
    return;
}

inline void split(int x, int y) {
    makeroot(x);
    access(y);
    splay(y);
    return;
}
inline void link(int x, int y) {
    split(x, y);
    si[f[x] = y] += s[x];
    push_up(y);
    return;
}
inline int get(int x) { return fa[x] == x ? x : fa[x] = get(fa[x]); }
inline int update(int x) {
    int l, r, ji = s[x] & 1, sum = s[x] >> 1, lsum = 0, rsum = 0, newp = inf, nl, nr;
    while (x) {
        push_down(x);
        nl = s[l = c[x][0]] + lsum;
        nr = s[r = c[x][1]] + rsum;
        if (nl <= sum && nr <= sum) {
            if (ji) {
                newp = x;
                break;
            } else if (newp > x)
                newp = x;
        }
        if (nl < nr)
            lsum += s[l] + si[x] + 1, x = r;
        else
            rsum += s[r] + si[x] + 1, x = l;
    }
    return splay(newp), newp;
}
signed main() {
    int n = read(), m = read(), x, y, z, ans = 0;
    for (int i = 1; i <= n; ++i) s[i] = 1, fa[i] = i, ans ^= i;
    while (m--) {
        string s;
        cin >> s;
        if (s == "A") {
            x = read();
            y = read();
            link(x, y);
            split(x = get(x), y = get(y));
            z = update(y);
            ans = ans ^ x ^ y ^ z;
            fa[x] = fa[y] = fa[z] = z;
        } else if (s == "Xor")
            printf("%lld\n", ans);
        else
            printf("%lld\n", get(read()));
    }
    return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }

来源：https://www.cnblogs.com/yspm/p/12375607.html