The meaning of problems
To a tree with the right side, there are two operations
- \ (C ~ ~ e_i W_i \) , the first \ (e_i \) side edges of the right to \ (W_i \) .
- \ (V_I Q ~ \) , from the query \ (V_I \) from the point furthest point.
analysis
Official explanations practice: Dynamic Maintenance diameter, and then ask support from two points, which can order + lca + Fenwick tree dfs. Dynamic Maintenance diameter can be dotted rule (dotted tree), which would be considered excessive and Treatment Center of the longest path, we only need to query respectively, each son of the root partition of the center, where the sub-tree of the longest chain from which to find the longest and times can be long, the star chart can be set to maintain. Each sub-tree may be used dfs + sequence segment tree maintenance. Complexity of O (nlog2n).
In fact, we do not consider the diameter, in order to maintain the same per son for the root partition of the center, where the sub-tree of the longest chain, can be deleted with a heap \ (ch [x] \) Maintenance \ (x \) for each son the longest chain of sub-tree query from \ (v \) when the point farthest distance, there are two cases
- \ (v \) point as a center partition, then the answer is \ (CH [v] \) .
- Jump \ (v \) dotted the tree ancestor, the first \ (ch [fa [v] ] \) is included in \ (v \) longest chain of sub-tree of deleting, the answer is \ (dis ( FA [V], V) + CH [FA [V]] \) , the \ (DIS \) may be used Fenwick tree + \ (LCA \) + DFS maintenance sequence.
All answers maximum value is the final answer. Chain Tree in each sub-sequence I used dfs + dynamic segment tree maintenance prescription, many of the details, because the Internet did not question this practice blog dotted tree, and he is just learning, many details of the deal are their own YY of .. the write very complex, feeling that only I can read (
Code
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define lson l,mid,p<<1
#define rson mid+1,r,p<<1|1
#define ll long long
using namespace std;
const int inf=1e9;
const int mod=1e9+7;
const int maxn=1e5+10;
typedef pair<int,int> pii;
int n,q;
int sz[maxn],mxp[maxn],vis[maxn],f[maxn],sum,rt;
ll tr[maxn];
int e[maxn];
int id[maxn];
vector<pii>g[maxn];
vector<int>son[maxn];
unordered_map<int,int>in[maxn],out[maxn],pd[maxn];
void add(int x,ll k){
while(x<=n) tr[x]+=k,x+=x&-x;
}
ll dor(int x){
ll ret=0;
while(x) ret+=tr[x],x-=x&-x;
return ret;
}
ll a[maxn*140],tag[maxn*140];
int ls[maxn*140],rs[maxn*140],rtt[maxn],tot;
void pdu(int p,ll k){a[p]+=k,tag[p]+=k;}
void up(int dl,int dr,int l,int r,int &p,ll k){
if(dl>dr||!dl) return;
a[++tot]=a[p],ls[tot]=ls[p],tag[tot]=tag[p],rs[tot]=rs[p],p=tot;
if(l==dl&&r==dr){
a[p]+=k;tag[p]+=k;
return;
}int mid=l+r>>1;
pdu(ls[p],tag[p]);pdu(rs[p],tag[p]);tag[p]=0;
if(dr<=mid) up(dl,dr,l,mid,ls[p],k);
else if(dl>mid) up(dl,dr,mid+1,r,rs[p],k);
else up(dl,mid,l,mid,ls[p],k),up(mid+1,dr,mid+1,r,rs[p],k);
a[p]=max(a[ls[p]],a[rs[p]]);
}
ll qy(int dl,int dr,int l,int r,int p){
if(l==dl&&r==dr) return a[p];
int mid=l+r>>1;
pdu(ls[p],tag[p]);pdu(rs[p],tag[p]);tag[p]=0;
if(dr<=mid) return qy(dl,dr,l,mid,ls[p]);
else if(dl>mid) return qy(dl,dr,mid+1,r,rs[p]);
else return max(qy(dl,mid,l,mid,ls[p]),qy(mid+1,dr,mid+1,r,rs[p]));
}
struct heap {
priority_queue<ll> A, B; // heap=A-B
void insert(ll x) { A.push(x); }
void erase(ll x) { B.push(x); }
ll top() {
while (!B.empty() && A.top() == B.top()) A.pop(), B.pop();
return A.top();
}
void pop() {
while (!B.empty() && A.top() == B.top()) A.pop(), B.pop();
A.pop();
}
ll top2() {
ll t = top(), ret;
pop();
ret = top();
A.push(t);
return ret;
}
int size() { return A.size() - B.size(); }
}ch[maxn];
struct LCA{
int sz[maxn],d[maxn],f[maxn],top[maxn],son[maxn],in[maxn],out[maxn],p[maxn],id[maxn],num;
ll dist[maxn];
void dfs1(int u){
sz[u]=1;d[u]=d[f[u]]+1;
for(pii x:g[u]){
if(x.fi==f[u]) continue;
f[x.fi]=u;dist[x.fi]=dist[u]+e[x.se],id[x.se]=x.fi;
dfs1(x.fi);
sz[u]+=sz[x.fi];
if(sz[x.fi]>sz[son[u]]) son[u]=x.fi;
}
}
void dfs2(int u,int t){
top[u]=t;in[u]=++num;p[num]=u;
if(son[u]) dfs2(son[u],t);
for(pii x:g[u]){
if(x.fi==f[u]||x.fi==son[u]) continue;
dfs2(x.fi,x.fi);
}
out[u]=num;
}
int lca(int x,int y){
while(top[x]!=top[y]){
if(d[top[x]]<d[top[y]]) swap(x,y);
x=f[top[x]];
}
if(d[x]<d[y]) swap(x,y);
return y;
}
}L;
ll dis(int x,int y){
return dor(L.in[x])+dor(L.in[y])-2*dor(L.in[L.lca(x,y)]);
}
void getrt(int u,int fa){
sz[u]=1;mxp[u]=0;
for(pii x:g[u]){
if(x.fi==fa||vis[x.fi]) continue;
getrt(x.fi,u);
sz[u]+=sz[x.fi];
mxp[u]=max(mxp[u],sz[x.fi]);
}
mxp[u]=max(mxp[u],sum-sz[u]);
if(mxp[u]<mxp[rt]) rt=u;
}
void calc(int u,int fa,int fart){
son[rt].pb(u);
in[rt][u]=son[rt].size();
if(fart!=-1) up(in[f[rt]][u],in[f[rt]][u],1,son[fart].size(),rtt[rt],dis(fart,u));
for(pii x:g[u]){
if(x.fi==fa||vis[x.fi]) continue;
pd[rt][x.se]=x.fi;id[x.se]=x.fi;
calc(x.fi,u,fart);
}
out[rt][u]=son[rt].size();
}
void solve(int u){
vis[u]=1;
for(pii x:g[u]){
if(vis[x.fi]) continue;
sum=sz[x.fi];mxp[rt=0]=inf;
getrt(x.fi,0);getrt(rt,0);
f[rt]=u;
calc(rt,0,u);
ch[u].insert(a[rtt[rt]]);
solve(rt);
}
ch[u].insert(0);
}
int main(){
//ios::sync_with_stdio(false);
//freopen("in","r",stdin);
scanf("%d",&n);
for(int i=1,x,y;i<n;i++){
scanf("%d%d%d",&x,&y,&e[i]);
g[x].pb(pii(y,i));g[y].pb(pii(x,i));
}
L.dfs1(1);L.dfs2(1,1);
for(int i=1;i<=n;i++) add(L.in[i],L.dist[i]),add(L.in[i]+1,-L.dist[i]);
sum=mxp[rt]=n;
getrt(1,0);calc(rt,0,-1);solve(rt);
scanf("%d",&q);
while(q--){
char c[2];
int x,j;
ll y;
scanf("%s",c);
if(c[0]=='C'){
scanf("%d%lld",&j,&y);
int gt=L.id[j];
ll ret=e[j];
x=id[j];
add(L.in[gt],y-ret),add(L.out[gt]+1,ret-y);
for(int i=x;f[i];i=f[i]){
ch[f[i]].erase(a[rtt[i]]);
up(in[f[i]][pd[f[i]][j]],out[f[i]][pd[f[i]][j]],1,son[f[i]].size(),rtt[i],y-ret);
ch[f[i]].insert(a[rtt[i]]);
}
e[j]=y;
}else{
scanf("%d",&x);
ll ans=ch[x].top();
for(int i=x;f[i];i=f[i]){
ch[f[i]].erase(a[rtt[i]]);
ans=max(ans,dis(f[i],x)+ch[f[i]].top());
ch[f[i]].insert(a[rtt[i]]);
}
printf("%lld\n",ans);
}
}
return 0;
}