树链剖分模板题（CodeForces - 343D Water Tree ）

题目链接
树剖本蒻苟一知没敢学，最后集训题目集出了这个题目，所以以此题学一下。
学懂后觉得不难，本人理解树链剖分本质上其实还是线段树，只不过通过对轻重链的划分，让树上节点连续化，以保留树上信息。从而使对树上链状的节点的修改连续化，使区间维护成为可能。可以去大佬博客学一下。

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int N = 1e6+50;
int dep[N], siz[N], fa[N], id[N], son[N], top[N];
int dfn[N], w[N];
int ver[N<<1], ne[N<<1], he[N];
int tot;
struct Node
{
    int l, r, val, add;
}tr[N<<1];
void add(int x, int y)
{
    ne[++tot] = he[x];
    ver[tot] = y;
    he[x] = tot;
}
void dfs1(int u, int f)
{
    fa[u] = f;
    dep[u] = dep[f]+1;
    siz[u] = 1;
    int mx = -1;
    for (int i = he[u]; i; i = ne[i])
    {
        int y = ver[i];
        if (y == f) continue;
        dfs1(y, u);
        siz[u] += siz[y];
        if (siz[y] > mx)
        {
            mx = siz[y];
            son[u] = y;
        }
    }
}
int cnt;
void dfs2(int u, int t)
{
    dfn[u] = ++cnt;
    top[u] = t;
    w[cnt] = 0;
    if (!son[u])
        return;
    dfs2(son[u], t);
    for (int i = he[u]; i; i = ne[i])
    {
        int v = ver[i];
        if (v == fa[u] || v == son[u])
            continue;
        dfs2(v, v);
    }
}
void build(int p, int l, int r)
{
    tr[p].val = 0;
    tr[p].add = -1;
    tr[p].l = l; tr[p].r = r;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(p<<1, l, mid);
    build(p<<1|1, mid+1, r);
}
void spread(int p)
{
    if (tr[p].add == -1) return;
    int l = p<<1, r = p<<1|1;
    tr[l].val = tr[p].add;
    tr[r].val = tr[p].add;
    tr[l].add = tr[r].add = tr[p].add;
    tr[p].add = -1;
}
int ask(int p, int k)
{
    if (tr[p].l == tr[p].r) return tr[p].val;
    spread(p);
    int mid = (tr[p].l+tr[p].r) >> 1;
    if (k <= mid) return ask(p<<1, k);
    else return ask(p<<1|1, k);
}
void change(int p, int l, int r, int v)
{
    if (l <= tr[p].l && tr[p].r <= r)
    {
        tr[p].val = v;
        tr[p].add = v;
        return;
    }
    spread(p);
    int mid = (tr[p].l + tr[p].r) >> 1;
    if (l <= mid) change(p<<1, l, r, v);
    if (r > mid) change(p<<1|1, l, r, v);
}
inline void cha(int x, int z)
{
    change(1, dfn[x], dfn[x]+siz[x]-1, z);
}
inline int qu(int x)
{
    return ask(1, dfn[x]);
}
void mc(int x, int z)
{
    int y = 1;
    while(top[x] != top[y])
    {
        if (dep[top[x]] < dep[top[y]])
            swap(x, y);
        change(1, dfn[top[x]], dfn[x], z);
        x = fa[top[x]];
    }
    if (dep[x] > dep[y])
        swap(x, y);
    change(1, dfn[x], dfn[y], z);
}
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i < n; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y);
        add(y, x);
    }
    dfs1(1, 0);
    dfs2(1, 0);
    build(1, 1, n);
    int q;
    scanf("%d", &q);
    while(q--)
    {
        int op;
        scanf("%d", &op);
        if (op ==1)
        {
            int k;
            scanf("%d", &k);
            cha(k, 1);
        }
        else if (op == 2)
        {
            int k;
            scanf("%d", &k);
            mc(k, 0);
        }
        else
        {
            int k;
            scanf("%d", &k);
            printf("%d\n", qu(k));
        }
    }
    return 0;
}