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LG传送门

表示乱搞比正解难想。

整体二分很好想吧。

但是为了好写快乐，我们选择三个\(\log\)的乱搞。

先树剖，线段树套堆维护区间最大值。对于一次修改，如果是插入，就把树上除了这条链的地方加上这个重要度，如果是删除则反之。注意线段树可以标记永久化，这里用的堆是一种（可能）叫懒惰堆的东西，直接看代码都能理解。

//written by newbiechd
#include <cstdio>
#include <cctype>
#include <vector>
#include <queue>
#include <algorithm>
#define R register
#define I inline
#define B 1000000
using namespace std;
const int N = 100003, M = 200003;
char buf[B], *p1, *p2;
I char gc() { return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, B, stdin), p1 == p2) ? EOF : *p1++; }
I int rd() {
    R int f = 0;
    R char c = gc();
    while (c < 48 || c > 57)
        c = gc();
    while (c > 47 && c < 58)
        f = f * 10 + (c ^ 48), c = gc();
    return f;
}
int s[N], fa[N], dep[N], siz[N], son[N], dfn[N], top[N], n, tim;
struct road {
    int x, y;
    road() {}
    road(int x, int y) : x(x), y(y) {}
}sta[N];
I int operator < (road a, road b) { return a.x ^ b.x ? a.x < b.x : a.y < b.y; }
struct event {
    int x, y, z;
}q[M];
struct prique {
    priority_queue <int> q1, q2;
    I void push(int x) { q1.push(x); }
    I void pop(int x) { q2.push(x); }
    I int top() {
        while (!q2.empty() && q1.top() == q2.top())
            q1.pop(), q2.pop();
        return q1.empty() ? -1 : q1.top();
    }
}e[N << 2];
vector <int> g[N];
I int max(int x, int y) { return x > y ? x : y; }
I void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
void dfs1(int x, int f) {
    fa[x] = f, dep[x] = dep[f] + 1, siz[x] = 1;
    for (R int i = 0, y, m = 0; i < s[x]; ++i)
        if ((y = g[x][i]) ^ f) {
            dfs1(y, x), siz[x] += siz[y];
            if (siz[y] > m)
                m = siz[y], son[x] = y;
        }
}
void dfs2(int x, int t) {
    dfn[x] = ++tim, top[x] = t;
    if (son[x])
        dfs2(son[x], t);
    for (R int i = 0, y; i < s[x]; ++i)
        if (!dfn[y = g[x][i]])
            dfs2(y, y);
}
void modify(int k, int l, int r, int x, int y, int z, int opt) {
    if (x == l && y == r) {
        opt ? e[k].pop(z) : e[k].push(z);
        return ;
    }
    R int p = k << 1, q = p | 1, m = (l + r) >> 1;
    if (y <= m)
        modify(p, l, m, x, y, z, opt);
    else
        if (m < x)
            modify(q, m + 1, r, x, y, z, opt);
        else
            modify(p, l, m, x, m, z, opt),
                modify(q, m + 1, r, m + 1, y, z, opt);
}
int query(int k, int l, int r, int x) {
    if (l == r)
        return e[k].top();
    R int p = k << 1, q = p | 1, m = (l + r) >> 1, o = e[k].top();
    if (x <= m)
        return max(o, query(p, l, m, x));
    else
        return max(o, query(q, m + 1, r, x));
}
I void insert(int x, int y, int z, int opt) {
    R int t = 0, i, l = 0;
    while (top[x] ^ top[y]) {
        if (dep[top[x]] < dep[top[y]])
            swap(x, y);
        sta[++t] = road(dfn[top[x]], dfn[x]), x = fa[top[x]];
    }
    if (dep[x] > dep[y])
        swap(x, y);
    sta[++t] = road(dfn[x], dfn[y]), sort(sta + 1, sta + t + 1);
    for (i = 1; i <= t; l = max(l, sta[i++].y))
        if (l + 1 < sta[i].x)
            modify(1, 1, n, l + 1, sta[i].x - 1, z, opt);
    if (l < n)
        modify(1, 1, n, l + 1, n, z, opt);
}
int main() {
    R int m, i, x, y, z, opt;
    n = rd(), m = rd();
    for (i = 1; i < n; ++i)
        x = rd(), y = rd(), g[x].push_back(y), g[y].push_back(x);
    for (i = 1; i <= n; ++i)
        s[i] = g[i].size();
    dfs1(1, 0), dfs2(1, 1);
    for (i = 1; i <= m; ++i) {
        opt = rd(), x = rd();
        if (opt == 0)
            y = rd(), z = rd(), q[i] = (event){x, y, z}, insert(x, y, z, opt);
        if (opt == 1)
            insert(q[x].x, q[x].y, q[x].z, opt);
        if (opt == 2)
            printf("%d\n", query(1, 1, n, dfn[x]));
    }
    return 0;
}

其实我觉得写正解的大佬们没必要diss我们这种只会乱搞的小蒟蒻啊这个乱搞的思路也挺精巧的