Query on a tree 树链剖分 [SPOJ-QTREE]

题目：https://www.spoj.com/problems/QTREE/en/

树链剖分模板题
注意一点，我们用dfs序对节点进行编号，其实也对应线段树中该节点与其父节点连边的编号
因此统计到最后时（两节点在同一条重链上）不能用在上面的节点的的编号，而是其重儿子的编号

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 2e5;

struct edge {
    int u,v,cost;
};
int fa[N],lv[N],siz[N],son[N],top[N],dfn[N],cnt=0;
vector<pii> G[N];
int n,seg[N<<2];
edge e[N];

int query(int s,int t,int l,int r,int p) {
    if(s<=l&&r<=t) return seg[p];
    int m=(l+r)/2;
    if(t<=m) return query(s,t,l,m,p<<1);
    if(m+1<=s) return query(s,t,m+1,r,p<<1|1);
    return max(query(s,t,l,m,p<<1),query(s,t,m+1,r,p<<1|1));
}

void update(int pos,int val,int l,int r,int p){
    if(l==r) {
        seg[p]=val;
        return;
    }
    int m=(l+r)/2;
    if(pos<=m) update(pos,val,l,m,p<<1);
    else update(pos,val,m+1,r,p<<1|1);
    seg[p]=max(seg[p<<1],seg[p<<1|1]);
}

int dfs1(int x,int fx) {
    son[x]=-1;
    siz[x]=1;
    fa[x]=fx;
    lv[x]=lv[fx]+1;
    for(auto p:G[x]) {
        int to = p.first;
        if(to==fx) continue;
        siz[x] += dfs1(to,x);
        if(son[x] == -1 || siz[to] > siz[son[x]]) son[x] = to;
    }
    return siz[x];
}

void dfs2(int x,int root) {
    top[x]=root;
    dfn[x]=++cnt;
    if(son[x]==-1) return;
    dfs2(son[x],root);
    for(auto p:G[x]) {
        int to = p.first;
        if(to == son[x] || to == fa[x]) continue;
        dfs2(to,to);
    }
}

int ask(int x,int y) {
    int ans = 0;
    while(top[x] != top[y]) {
        if(lv[top[x]]<lv[top[y]]) swap(x,y);
        ans=max(ans,query(dfn[top[x]],dfn[x],1,n,1));
        x=fa[top[x]];
    }
    if(dfn[x] != dfn[y]) {
        if(dfn[x]>dfn[y]) swap(x,y);
        ans=max(ans,query(dfn[son[x]],dfn[y],1,n,1));
    }
    return ans;
}

int a,b,c;
string o;

void solve() {
    memset(seg,0,sizeof(seg));
    cnt = 0;
//    cin>>n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) G[i].clear();
    for(int i=1;i<n;i++){
        scanf("%d %d %d",&a,&b,&c);
        G[a].push_back({b,c});
        G[b].push_back({a,c});
        e[i]={a,b,c};
    }
    dfs1(1,-1);
    dfs2(1,1);
    for(int i=1;i<n;i++){
        if(lv[e[i].u] > lv[e[i].v]) swap(e[i].u, e[i].v);
        update(dfn[e[i].v], e[i].cost, 1,n,1);
    }
    char str[10];
    while(true) {
        scanf(" %s",str);
        if(str[0] == 'Q') {
//            cin>>a>>b;
            scanf("%d %d",&a,&b);
//            cout<<ask(a,b)<<'\n';
            printf("%d\n",ask(a,b));
        } else if(str[0] == 'C') {
//            cin>>a>>b;
            scanf("%d %d",&a,&b);
            update(dfn[e[a].v], b, 1,n,1);
        } else break;
    }
}

int main(){
    ios::sync_with_stdio(false);
    int t;
    scanf("%d",&t);
    while(t--) solve();
    return 0;
}