树链剖分板子

const int N = 1e5 + 5;
vector<int> g[N];
int fa[N], dp[N], sz[N], son[N], top[N], dfn[N], to[N], cnt = 0, n;
void dfs1(int u, int o) {
    fa[u] = o;
    sz[u] = 1;
    dp[u] = dp[o] + 1;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i];
        if(v != o) {
            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;
    dfn[u] = ++cnt;
    to[cnt] = u;
    if(!son[u]) return ;
    dfs2(son[u], t);
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i];
        if(v != fa[u] && v != son[u]) dfs2(v, v);
    }
}
void add(int u, int v, int x) {
    int fu = top[u], fv = top[v];
    while(fu != fv) {
        if(dp[fu] >= dp[fv]) update(dfn[fu], dfn[u], x, 1, 1, n), u = fa[fu], fu = top[u];
        else update(dfn[fv], dfn[v], x, 1, 1, n), v = fa[fv], fv = top[v];
    }
    if(dfn[u] <= dfn[v]) update(dfn[u], dfn[v], x, 1, 1, n);
    else update(dfn[v], dfn[u], x, 1, 1, n);
}
int sum(int u, int v) {
    int ans = 0, fu = top[u], fv = top[v];
    while(fu != fv) {
        if(dp[fu] >= dp[fv]) ans += query(dfn[fu], dfn[u], 1, 1, n), u = fa[fu], fu = top[u];
        else ans += query(dfn[fv], dfn[v], 1, 1, n), v = fa[fv], fv = top[v];
    }
    if(dfn[u] <= dfn[v]) ans += query(dfn[u], dfn[v], 1, 1, n);
    else ans += query(dfn[v], dfn[u], 1, 1, n);
    return ans;
}
int lca(int u, int v) {
    int fu = top[u], fv = top[v];
    while(fu != fv) {
        if(dp[fu] >= dp[fv]) u = fa[fu], fu = top[u];
        else v = fa[fv], fv = top[v];
    }
    if(dp[u] <= dp[v]) return u;
    else v;
}