The meaning of problems: the tree given any two points, and a maximum value on the request path, a modify operation with a single point
Tree chain split ideas:
1, the tree depth and parent node dfs obtained point, then find the severity of his son (heavy son is the point of most of the sub-tree, the rest are all sons of light), with a son pointing to an array of heavy son of each node
2, the tree a second time dfs, for all heavy son, obtained his top node is the son along each weight heavy chain can reach the farthest of the ancestors, then dfs maintain order, each node records access order and the first few visits of which node, the son of the top node is the light itself
Then we get
dfs Sequence: 1,491,314,810,372,611,125
top array: 1,111,181,033,222,125 (sequence corresponding to dfs)
So we put the tree demolition chain (top value compared to the same point on a chain) became a section of the maintenance order with the dfs tree line, you can quickly obtain the maximum value and the value chain and the,
For any two points, we just need to find the answer to all the chain turn on the path then be merged, to prove the number of light and heavy chain on the path of no more than logn, so a single query complexity is O ((logn ) ^ 2)
The total time complexity O (q (logn) ^ 2)
In the calculation of two answers when taking a smart way. If the first two points not on the same chain, we always make a greater depth of the point x to jump up top [x], and the statistics answer this chain until the two points on the same chain, Finally the answer to when calculated on a chain
AC Code (template)
#include<bits/stdc++.h> using namespace std; const int N = 1e5+5; struct Edge { int v,next; }edge[N<<1]; int sum[N<<2],mx[N<<2],n; int head[N],tot,dep[N],fa[N],sz[N],son[N],top[N],id[N],rk[N],cnt,val[N]; void init() { memset(head,-1, sizeof(head)); tot=0; } void add(int u,int v) { edge[tot].v=v; edge[tot].next=head[u]; head[u]=tot++; } void dfs1(int u,int f) { dep[u]=dep[f]+1; fa[u]=f; sz[u]=1; for(int i=head[u];~i;i=edge[i].next) { int v=edge[i].v; if(v==f)continue; dfs1(v,u); sz[u]+=sz[v]; if(sz[v]>sz[son[u]])son[u]=v; } } void dfs2(int u,int t) { top[u]=t; id[u]=++cnt; rk[cnt]=u; if(!son[u])return; dfs2(son[u],t); for(int i=head[u];~i;i=edge[i].next) { int v=edge[i].v; if(v!=son[u]&&v!=fa[u])dfs2(v,v); } } void pushup(int rt) { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; mx[rt]=max(mx[rt<<1],mx[rt<<1|1]); } void build(int l,int r,int rt) { if(l==r) { mx[rt]=sum[rt]=val[rk[l]]; return; } int m=(l+r)>>1; build(l,m,rt<<1); build(m+1,r,rt<<1|1); pushup(rt); } int querySum(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R)return sum[rt]; int m=(l+r)>>1; int res=0; if(L<=m)res+=querySum(L,R,l,m,rt<<1); if(R>m)res+=querySum(L,R,m+1,r,rt<<1|1); return res; } int queryMax(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R)return mx[rt]; int m=(l+r)>>1; int res=-1e9; if(L<=m)res=max(res,queryMax(L,R,l,m,rt<<1)); if(R>m)res=max(res,queryMax(L,R,m+1,r,rt<<1|1)); return res; } void update(int pos,int val,int l,int r,int rt) { if(l==r){ sum[rt]=mx[rt]=val; return; } int m=(l+r)>>1; if(pos<=m)update(pos,val,l,m,rt<<1); else update(pos,val,m+1,r,rt<<1|1); pushup(rt); } int getSum(int x,int y) { int ans=0; while(top[x]!=top[y]) { if(dep[top[x]]<dep[top[y]])swap(x,y); ans+=querySum(id[top[x]],id[x],1,n,1); x=fa[top[x]]; } if(id[x]>id[y])swap(x,y); ans+=querySum(id[x],id[y],1,n,1); return ans; } int getMax(int x,int y) { int ans=-1e9; while (top[x]!=top[y]) { if(dep[top[x]]<dep[top[y]])swap(x,y); ans=max(ans,queryMax(id[top[x]],id[x],1,n,1)); x=fa[top[x]]; } if(id[x]>id[y])swap(x,y); ans=max(ans,queryMax(id[x],id[y],1,n,1)); return ans; } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); init(); int u,v; cin>>n; for(int i=1;i<=n-1;i++) { cin>>u>>v; add(u,v); add(v,u); } for(int i=1;i<=n;i++)cin>>val[i]; dfs1(1,1); dfs2(1,1); build(1,n,1); int q,x,y; string op; cin>>q; while(q--) { cin>>op>>x>>y; if(op=="QMAX")cout<<getMax(x,y)<<'\n'; else if(op=="QSUM")cout<<getSum(x,y)<<'\n'; else update(id[x],y,1,n,1); } return 0; }