## Hello everyone , I really like violent data structures , So I went through the problem with block trees

subject ：

One tree has n Nodes , The numbers are 1 To n, Each node has a weight w.

I. CHANGE u t : Put the node u The weight of is changed to t

II. QMAX u v: Ask from point u point-to-point v The maximum weight of the node on the path of

III. QSUM u v: Ask from point u point-to-point v The weight sum of nodes on the path of

Be careful ： From the point of u point-to-point v The nodes on the path of include u and v In itself

We can roughly divide the tree into $$\sqrt{n}$$ block , The length of the path to the root node in each block and the maximum point weight are maintained , and , obviously , We can do this by looking for their $$LCA$$ To find out about their path , And here we've partitioned the trees .

So the worst jump time complexity in the same block is $$O(\sqrt{n})$$

The worst time complexity of jumping between blocks is $$O(\sqrt{n})$$

Relaxed AC This topic

There are more detailed comments in the code , Post code

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
struct cc{
int to,nex;
}e[maxn],dis[maxn];
void add1(int u,int v)// On the edge of the original tree
{
++cnt1;
e[cnt1].to=v;
}
void add2(int u,int v)// After partition, inside the block, beside the tree
{
++cnt2;
dis[cnt2].to=v;
dis[cnt2].nex=h[u];
h[u]=cnt2;
}
int rt[maxn],mx[maxn],sum[maxn],siz[maxn];
int n,m,v[maxn],deep[maxn],len,fa[maxn];
void dfs(int u,int f,int dep)
{
deep[u]=dep;
int tmp=rt[u];
fa[u]=f;
{
int v=e[i].to;
if(v!=f)
{
if(siz[tmp]+1<len)
{
add2(u,v);// The trees in the block are connected to the sides
rt[v]=tmp;
++siz[tmp];
}
dfs(v,u,dep+1);
}
}
}
void build(int u,int num,int vmx)// Maintain the current node , The sum to the root node in the block , Maximum
{
num+=v[u],sum[u]=num;
vmx=max(vmx,v[u]),mx[u]=vmx;
for(int i=h[u];i;i=dis[i].nex)
build(dis[i].to,num,vmx);
}
int query(int a,int b,int tag)
{
int ans1=0;//QSUM
int ans2=-(1<<30);//QMAX
while(a!=b)// It's like doubling , It's just that the distance here is sqrt(n)
{
if(deep[a]<deep[b]) swap(a,b);
if(rt[a]==rt[b])// If they belong to the same block
{
ans1+=v[a];
ans2=max(ans2,v[a]);
a=fa[a];// Because in the same block , The complexity of violent jumping is just O(sqrt(n))
}
else
{
if(deep[rt[a]]<deep[rt[b]]) swap(a,b);// The depth of the block is deeper
ans1+=sum[a];
ans2=max(ans2,mx[a]);
a=fa[rt[a]];// Just jump one block
}
}
ans1+=v[a];
ans2=max(ans2,v[a]);// Update their LCA Value
if(tag==0) return ans2;
else return ans1;
}
void change(int u,int x)
{
v[u]=x;
if(u==rt[u]) build(u,0,-(1<<30));// If it is a root node in a block, the whole block is updated
else build(u,sum[fa[u]],mx[fa[u]]);// If not , It started with his father
}
int main()
{
int x,y;
scanf("%d",&n);
len=sqrt(n);
for(int i=1;i<n;++i)
for(int i=1;i<=n;++i)
scanf("%d",&v[i]),rt[i]=i;
dfs(1,0,0);
for(int i=1;i<=n;++i)
if(rt[i]==i)
build(i,0,-(1<<30));
scanf("%d",&m);
char opt[30];
for(int i=1;i<=m;++i)
{
scanf("%s%d%d",opt,&x,&y);
if(opt[1]=='M')//QMAX
printf("%d\n",query(x,y,0));//01 Maintenance questions
else if(opt[1]=='S')//QSUM
printf("%d\n",query(x,y,1));//01 Maintenance questions
else //CHANGE
change(x,y);
}
return 0;
}

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