Jamie and Tree CodeForces

大意: n节点树, 每个点有权值, 三种操作: 1,换根. 2, lca(u,v)的子树权值全部增加x. 3, 查询子树权值和.

先不考虑换根, 考虑子树x加v的贡献

(1)对fa[x]到根的树链贡献为sz[x]*v;

(2)对x子树内的点y贡献为sz[y]*v;

步骤(1)可以用单点更新子树求和实现, 步骤(2)可以子树更新单点求和实现

然后就是换根板子题了.

感觉蠢得不行的题啊, 还是打了好久, 怎么能这么菜啊

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) {REP(i,1,n) cout<<a[i]<<' ';hr;}
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif


int n, a[N];
vector<int> g[N];
int sz[N], dep[N], fa[N], L[N], R[N];
int son[N], top[N];
ll c[N], c2[N];
void add(int x, ll v) {
	for (; x<=n; x+=x&-x) c[x]+=v;
}
void add2(int x, int v) {
	for (; x; x^=x&-x) c2[x]+=v;
}
void add2(int l, int r, int v) {
	add2(r,v),add2(l-1,-v);
}
ll qry(int x) {
	ll ret = 0;
	for (; x; x^=x&-x) ret += c[x];
	return ret;
}
ll qry(int l, int r) {
	return qry(r)-qry(l-1);
}
ll qry2(int x) {
	ll ret = 0;
	for (; x<=n; x+=x&-x) ret += c2[x];
	return ret;
}
void dfs(int x, int d, int f) {
	sz[x] = 1, dep[x]=d, fa[x]=f, L[x]=++*L, add(L[x],a[x]);
	for (int y:g[x]) if (y!=f) {
		dfs(y,d+1,x); sz[x]+=sz[y];
		if (sz[y]>sz[son[x]]) son[x]=y;
	}
	R[x]=*L;
}
void dfs2(int x, int tf) {
	top[x]=tf;
	if (son[x]) dfs2(son[x],tf);
	for (int y:g[x]) if (y!=fa[x]&&y!=son[x]) dfs2(y,y);
}
int lca(int x, int y) {
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		x = fa[top[x]];
	}
	if (dep[x]>dep[y]) swap(x,y);
	return x;
}
int lca(int u, int v, int rt) {
	int L = lca(u,v);
	if (lca(rt,L)!=L) return L;
	int x=lca(u,rt),y=lca(v,rt);
	if (dep[x]<dep[y]) swap(x,y);
	return x;
}
int calc(int x, int y) {
	int f = x, pre = 0, lca;
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		pre = top[x], x = fa[pre];
	}
	if (x==y) lca=x;
	else lca=dep[x]<dep[y]?x:y;
	if (lca!=f) return 0;
	return fa[pre]==f?pre:son[f];
}
void update(int x, int v) {
	add(L[fa[x]],(ll)sz[x]*v);
	add2(L[x],R[x],v);
}
void update(int x, int v, int rt) {
	if (x==rt) return add2(1,n,v);
	int t = calc(x,rt);
	if (!t) return update(x,v);
	add2(1,n,v),update(t,-v);
}
ll query(int x) {
	return qry(L[x],R[x])+sz[x]*qry2(L[x]);
}
ll query(int x, int rt) {
	if (x==rt) return query(1);
	int t = calc(x,rt);
	if (!t) return query(x);
	return query(1)-query(t);
}

int main() {
	int q, rt = 1;
	scanf("%d%d", &n, &q);
	REP(i,1,n) scanf("%d", a+i);
	REP(i,2,n) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0,0),dfs2(1,1);
	while (q--) {
		int op, u, v, w;
		scanf("%d%d", &op, &u);
		if (op==1) rt=u;
		else if (op==2) {
			scanf("%d%d", &v, &w);
			update(lca(u,v,rt),w,rt);
		}
		else printf("%lld
", query(u,rt));
	}
}