有个换根的操作十分麻烦,于是考虑并不真正地换根。

考虑以 11 为根建树,如果你目前被换成的根为 rtrt ,那对于点对 (u,v)(u,v)lcalca 是什么。

手玩一下不难发现是 lca(u,rt),lca(u,v),lca(v,rt)lca(u, rt), lca(u, v), lca(v, rt) 中深度较为深的那个。

那么现在我们要处理的问题就变成了维护 lcalca 的子树和。

维护子树和,是 dfndfn 的一个经典运用,考虑这题,发现大力讨论就可以做了。

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#include <bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &a){
T w = 1; a = 0;
char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') w = -1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) a = (a * 10) + (ch - '0');
a *= w;
}
template <typename T> inline void ckmax(T &a, T b){a = a > b ? a : b;}
template <typename T> inline void ckmin(T &a, T b){a = a < b ? a : b;}
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mii map<int, int>
#define pii pair<int, int>
#define vi vector<int>
#define si set<int>
#define ins insert
#define era erase
#define Debug(x) cout << #x << " = " << x << endl
#define For(i,l,r) for (int i = l; i <= r; ++i)
#define foR(i,l,r) for (int i = l; i >= r; --i)
#define int long long
const int N = 1e5 + 10;
int n, m;
int val[N];
vi to[N];
int dep[N], dp[N][32];
int dfn[N], dfntot, rnk[N];
int low[N];
void dfs (int u, int fa) {
rnk[dfn[u] = ++dfntot] = u;
dep[u] = dep[fa] + 1;
dp[u][0] = fa;
for (int i = 1; (1 << i) <= dep[u]; ++i)
dp[u][i] = dp[dp[u][i - 1]][i - 1];
for (auto v : to[u]) {
if (v == fa) continue;
dfs(v, u);
}
low[u] = dfntot;
}
int LCA (int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
foR (i, 20, 0) if (dep[x] - (1 << i) >= dep[y]) x = dp[x][i];
if (x == y) return x;
foR (i, 20, 0) if (dp[x][i] != dp[y][i]) x = dp[x][i], y = dp[y][i];
return dp[x][0];
}
int LCA_son (int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
foR (i, 20, 0) if (dep[x] - (1 << i) > dep[y]) x = dp[x][i];
if (dp[x][0] == y) return x;
foR (i, 20, 0) if (dp[x][i] != dp[y][i]) x = dp[x][i], y = dp[y][i];
return x;
}

#define ls x << 1
#define rs x << 1 | 1

int tag[N << 2], sum[N << 2];
inline void pushup (int x) { sum[x] = sum[ls] + sum[rs]; }
inline void push (int x, int l, int r, int v) {
tag[x] += v, sum[x] += (r - l + 1) * v;
}
inline void pushdown (int x, int l, int r) {
if (tag[x]) {
int mid = l + r >> 1;
push(ls, l, mid, tag[x]), push(rs, mid + 1, r, tag[x]);
tag[x] = 0;
}
}
void build (int x, int l, int r) {
if (l == r) return (sum[x] = val[rnk[l]]), void();
int mid = l + r >> 1;
build (ls, l, mid), build (rs, mid + 1, r);
pushup(x);
}

void update (int x, int l, int r, int ll, int rr, int v) {
if (ll <= l && r <= rr) return push(x, l, r, v), void();
int mid = l + r >> 1;
pushdown(x, l, r);
if (ll <= mid) update (ls, l, mid, ll, rr, v);
if (rr > mid) update(rs, mid + 1, r, ll, rr, v);
pushup(x);
}

int query (int x, int l, int r, int ll, int rr) {
if (ll <= l && r <= rr) return sum[x];
int ans = 0, mid = l + r >> 1;
pushdown(x, l, r);
if (ll <= mid) ans += query (ls, l, mid, ll, rr);
if (rr > mid) ans += query (rs, mid + 1, r, ll, rr);
return ans;
}
int rt, Mck;
signed main() {
read(n), read(m);
For (i, 1, n) read(val[i]);
For (i, 2, n) {
int u, v; read(u), read(v);
to[u].pb(v), to[v].pb(u);
}
dfs(1, 0);
build (1, 1, n);
rt = 1;
For (i, 1, m) {
int op; read(op);
if (op == 1) read(rt);
if (op == 2) {
int u, v, det; read(u), read(v); read(det);
int lca = LCA(u, v);
int tmp = LCA(u, rt);
(dep[tmp] > dep[lca]) ? lca = tmp : Mck = 1;
tmp = LCA(v, rt);
(dep[tmp] > dep[lca]) ? lca = tmp : Mck = 1;
int llccaa = LCA(lca, rt);
if (llccaa != lca && llccaa != rt) {
update(1, 1, n, dfn[lca], low[lca], det);
continue;
}
if (lca == rt) push(1, 1, n, det);
else if (llccaa == rt) {
update (1, 1, n, dfn[lca], low[lca], det);
}
else {
Mck = LCA_son(lca, rt);
push(1, 1, n, det);
update(1, 1, n, dfn[Mck], low[Mck], -det);
}
}
if (op == 3) {
int u; read(u);
if (u == rt) {
printf ("%lld\n", query(1, 1, n, 1, n));
continue; }
int lca = LCA(u, rt);
if (lca != u && lca != rt) {
printf ("%lld\n", query(1, 1, n, dfn[u], low[u]));
continue;
}
if (u == rt) printf ("%lld\n", query (1, 1, n, 1, n));
else if ( lca == rt )
printf ("%lld\n", query(1, 1, n, dfn[u], low[u]));
else {
Mck = LCA_son(u, rt);
printf ("%lld\n", query (1, 1, n, 1, n) - query (1, 1, n, dfn[Mck], low[Mck]));
}
}
}
return 0;
}