2019 icpc上海網絡賽A（線段樹動態最遠點）

題目鏈接：https://nanti.jisuanke.com/t/41398

思路：類似與線段樹維護直徑的方法具體可以看這篇博客

https://www.cnblogs.com/TinyWong/p/11260601.html

#pragma GCC optimize(3)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-8
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int ls[100008],rs[100008],qw,n,w[100008],a[100008],b[100008],q,x,y;
char op[2];
ll ans;
vector<P>g[100008];
struct as{
    int l,r;
    ll m1,m2,m3,m4,lz;
    inline as operator+(const as&b)const
    {
        as c;c.m1=max(m1,b.m1);c.m2=max(m2,b.m2);
        c.m3=max(max(m3,b.m3),m2+b.m1);
        c.m4=max(max(m4,b.m4),m1+b.m2);
        c.l=l,c.r=b.r;c.lz=0;
        return c;
    }
}tr[1000008<<2];
void add(int d,ll pos)
{
    tr[d].m1+=pos;
    tr[d].m2-=2ll*pos;
    tr[d].m3-=pos;
    tr[d].m4-=pos;
    tr[d].lz+=pos;
}
void pushh(int d)
{
    if(tr[d].lz==0) return;
    add(lc,tr[d].lz);
    add(rc,tr[d].lz);
    tr[d].lz=0;
}
void build(int d,int l,int r)
{
    tr[d].l=l,tr[d].r=r;tr[d].lz=0;
    if(l==r) {tr[d].m1=tr[d].m2=tr[d].m3=tr[d].m4=0;return;}
    int mid=(l+r)>>1;
    build(lc,l,mid);
    build(rc,mid+1,r);
    tr[d]=tr[lc]+tr[rc];
}
void add(int d,int l,int r,int pos)
{
    if(tr[d].l==l&&tr[d].r==r) {add(d,pos);return;}
    pushh(d);
    int mid=(tr[d].l+tr[d].r)>>1;
    if(mid>=r)  add(lc,l,r,pos);
    else if(l>mid)  add(rc,l,r,pos);
    else  add(lc,l,mid,pos),add(rc,mid+1,r,pos);
    tr[d]=tr[lc]+tr[rc];
}
as query(int d,int l,int r)
{
    if(tr[d].l==l&&tr[d].r==r)  {return tr[d];}
    pushh(d);
    int mid=(tr[d].l+tr[d].r)>>1;
    if(mid>=r)  return query(lc,l,r);
    else if(l>mid)  return query(rc,l,r);
    else  return query(lc,l,mid)+query(rc,mid+1,r);
}
void dfs(int u,int fa)
{
    ls[u]=++qw;
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i].first;
        if(v==fa) continue;
        dfs(v,u);++qw;
    }
    rs[u]=qw;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    FOR(i,1,n-1)
    {
        si(a[i]),si(b[i]),si(w[i]);
        g[a[i]].pb(P(b[i],w[i]));
        g[b[i]].pb(P(a[i],w[i]));
    }
    dfs(1,0);
    build(1,1,qw);
    FOR(i,1,n-1)
    {
        if(ls[a[i]]>ls[b[i]]) swap(a[i],b[i]);
        add(1,ls[b[i]],rs[b[i]],w[i]);
    }
    cin>>q;
    while(q--)
    {
        ss(op);
        if(op[0]=='C')
        {
            si(x),si(y);
            add(1,ls[b[x]],rs[b[x]],y-w[x]);
            w[x]=y;
        }
        else
        {
            si(x);
            as ans1=query(1,1,ls[x]),ans2=query(1,ls[x],qw),ans3=query(1,ls[x],ls[x]);
            ans=max(ans1.m4,ans2.m3);
            ans+=ans3.m1;
            printf("%lld\n",ans);
        }
    }
    return 0;
}