给一个 $n$ 个点的带权树，每次修改一个点的权值，或者询问到 $x$ 距离不超过 $k$ 的点的权值和，强制在线
sol：
套路题，首先搞出一个点分树，每个重心，以到重心的距离为下标，点权为权值建两棵线段树，一个用来统计答案，一个用来消除对父节点的影响
每次修改和讯询问都是暴力爬树高，在经过的每棵线段树里修改就可以了
好了，然后我们来 $D$ 这个 sb 博主
道理我都懂，你求 LCA 为什么要树剖啊，复杂度强行加 log
明明树状数组就可以干的事情（单点加，区间和）你为什么要线段树啊，强行多大常数
OYJason 8160ms 就能过的题强行被我搞到了 13796ms
我真是个 sb
复杂度大概是 $O(nlogn + mlog^2n)$ ?
#include<bits/stdc++.h> #define LL long long using namespace std; inline int read() { int x = 0,f = 1;char ch = getchar(); for(;!isdigit(ch);ch = getchar())if(ch == '-')f = -f; for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0'; return x * f; } const int maxn = 500010; int n,m,lastans; int first[maxn],to[maxn << 1],nx[maxn << 1],cnt; inline void add(int u,int v){to[++cnt] = v;nx[cnt] = first[u];first[u] = cnt;} inline void ins(int u,int v){add(u,v);add(v,u);} namespace LCA //RMQ is not good ? { int dep[maxn],bl[maxn],fa[maxn],size[maxn]; inline void dfs1(int x) { size[x] = 1; for(int i=first[x];i;i=nx[i]) { if(to[i] == fa[x])continue; fa[to[i]] = x;dep[to[i]] = dep[x] + 1; dfs1(to[i]);size[x] += size[to[i]]; } } inline void dfs2(int x,int col) { int k = 0; bl[x] = col; for(int i=first[x];i;i=nx[i]) if(dep[to[i]] > dep[x] && size[to[i]] > size[k])k = to[i]; if(!k)return; dfs2(k,col); for(int i=first[x];i;i=nx[i]) if(dep[to[i]] > dep[x] && to[i] != k)dfs2(to[i],to[i]); } inline int lca(int x,int y) { while(bl[x] != bl[y]) { if(dep[bl[x]] < dep[bl[y]])swap(x,y); x = fa[bl[x]]; } return dep[x] > dep[y] ? y : x; } } namespace SEG { int root[maxn][2],ls[maxn << 4],rs[maxn << 4],val[maxn << 4],size; inline void Insert(int &x,int l,int r,int pos,int va) { if(!x) x = ++size; val[x] += va; if(l == r)return; int mid = (l + r) >> 1; if(pos <= mid)Insert(ls[x],l,mid,pos,va); else Insert(rs[x],mid + 1,r,pos,va); } inline int query(int x,int l,int r,int L,int R) { if(!x)return 0; if(L <= l && r <= R)return val[x]; int mid = (l + r) >> 1,ans = 0; if(L <= mid)ans += query(ls[x],l,mid,L,R); if(R > mid)ans += query(rs[x],mid + 1,r,L,R); return ans; } } inline int q_dis(int x,int y){return LCA::dep[x] + LCA::dep[y] - (LCA::dep[LCA::lca(x,y)] << 1);} int size[maxn],f[maxn],vis[maxn],d[maxn],v[maxn],son,root; int nfa[maxn]; inline void solve(int rt,int type,int x,int fa) { SEG::Insert(SEG::root[rt][type],0,n,d[x],v[x]); for(int i=first[x];i;i=nx[i]) { if(to[i] == fa || vis[to[i]])continue; d[to[i]] = d[x] + 1; solve(rt,type,to[i],x); } } inline void getroot(int x,int fa) { size[x] = 1,f[x] = 0; for(int i=first[x];i;i=nx[i]) { if(to[i] == fa || vis[to[i]])continue; getroot(to[i],x);size[x] += size[to[i]]; f[x] = max(f[x],size[to[i]]); }f[x] = max(f[x],son - size[x]); if(f[x] < f[root])root = x; } inline void work(int x) { vis[x] = 1;d[x] = 0;solve(x,0,x,0); for(int i=first[x];i;i=nx[i]) { if(vis[to[i]])continue; son = size[to[i]];root = 0; d[to[i]] = 1;getroot(to[i],0); solve(root,1,to[i],x); nfa[root] = x;work(root); } } inline int query(int x,int k) { int ret = SEG::query(SEG::root[x][0],0,n,0,k); for(int i=x;nfa[i];i = nfa[i]) { int du = q_dis(x,nfa[i]); ret += SEG::query(SEG::root[nfa[i]][0],0,n,0,k - du); ret -= SEG::query(SEG::root[i][1],0,n,0,k - du); }return ret; } inline void update(int x,int k) { int delt = k - SEG::query(SEG::root[x][0],0,n,0,0); SEG::Insert(SEG::root[x][0],0,n,0,delt); for(int i=x;nfa[i];i = nfa[i]) { int du = q_dis(x,nfa[i]); SEG::Insert(SEG::root[nfa[i]][0],0,n,du,delt); SEG::Insert(SEG::root[i][1],0,n,du,delt); } } int main() { //freopen("1.in","r",stdin); //freopen("1w.out","w",stdout); n = read(),m = read(); for(int i=1;i<=n;i++)v[i] = read(); for(int i=2;i<=n;i++) { int u = read(),v = read(); ins(u,v); }LCA::dfs1(1);LCA::dfs2(1,1); f[0] = son = n; getroot(1,0);work(root); while(m--) { int opt = read(),x = read() ^ lastans,y = read() ^ lastans; if(opt)update(x,y); else printf("%d ",lastans = query(x,y)); } }