CF343D Water Tree（水树？）

题面

思路：树剖+线段树。

树剖的裸题。

线段树维护时标记下放的过程中，有个小细节需要注意：在父亲节点的标记下放完成后，标记为\(-1\)，因为如果置为\(0\)，那么在下次操作时会把\(0\)当做你的操作下放下去....啊可能也有别的处理方法，反正我是这么处理的qwq。

代码：

#include <bits/stdc++.h>
using namespace std;

template<typename temp>void read(temp &x){
    x = 0;temp f = 1;char ch;
    while(!isdigit(ch = getchar())) (ch == '-') and (f = -1);
    for(x = ch^48; isdigit(ch = getchar()); x = (x<<1)+(x<<3)+(ch^48));
    x *= f;
}
template <typename temp, typename ...Args>void read(temp& a, Args& ...args){read(a), read(args...);}

const int maxn = 1e6;

int n, m, ans;

vector<int> v[maxn];

struct poutree{
    #define ls (now<<1)
    #define rs (now<<1|1)
    #define mid ((l+r)>>1)
    int cnt, dfn[maxn], low[maxn], height_son[maxn], size[maxn], top[maxn], fa[maxn], dep[maxn];
    struct no{
        int l, r, tag;
    }node[maxn<<1];
    void build_poutree(int now){
        size[now] = 1;
        for(int i = 0; i < v[now].size(); i ++){
            int to = v[now][i];
            if(dep[to]) continue;
            dep[to] = dep[fa[to]=now] + 1;
            build_poutree(to);
            size[now] += size[to];
            if(size[to] > size[height_son[now]]) height_son[now] = to;
        }
    }
    void dfs(int now, int topfa){
        top[now] = topfa, dfn[now] = ++cnt;
        if(height_son[now]) dfs(height_son[now],topfa);
        for(int i = 0; i < v[now].size(); i ++){
            int to = v[now][i];
            if(fa[now] == to or height_son[now] == to) continue;
            dfs(to,to);
        }
    }
    void down(int now){
        if(node[now].tag == -1) return;
        node[ls].tag = node[rs].tag = node[now].tag;
        node[now].tag = -1;
    }
    void build_segtree(int l, int r, int now){
        node[now].l = l, node[now].r = r;
        if(l == r) return(void)(node[now].tag = 0);
        build_segtree(l, mid, ls), build_segtree(mid+1, r, rs);
        return;
    }
    void chenge(int l, int r, int now, int val){
        if(node[now].l > r or node[now].r < l) return;
        if(l <= node[now].l and node[now].r <= r) return(void)(node[now].tag = val);
        down(now);
        chenge(l, r, ls, val), chenge(l, r, rs, val);
        return;
    }
    void quary(int l, int r, int now, int &val){
        if(node[now].l > r or node[now].r < l) return;
        if(l <= node[now].l and node[now].r <= r) return(void)(val += node[now].tag);
        down(now);
        quary(l, r, ls, val), quary(l, r, rs, val);
        return;
    }
    void function(int opt, int x){
        if(opt == 1) chenge(dfn[x], dfn[x]+size[x]-1, 1, 1);
        if(opt == 2) while(x) chenge(dfn[top[x]], dfn[x], 1, 0), x = fa[top[x]];
        if(opt == 3) quary(dfn[x], dfn[x], 1, ans=0), printf("%d\n", ans);
    }
}tree;

signed main(){
    read(n);
    for(int i = 1, x, y; i < n; i ++){
        read(x, y);
        v[x].push_back(y), v[y].push_back(x);
    }
    tree.build_poutree(1*(tree.dep[1]=1)), tree.dfs(1,1), tree.build_segtree(1,n,1);
    read(m);
    for(int opt, x; m; m --){
        read(opt, x);
        tree.function(opt, x);
    }
    return 0;
}