[CF334D]Water Tree

【原题】

Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Each vertex is a reservoir which can be either empty or filled with water.

The vertices of the tree are numbered from 1 to n with the root at vertex 1. For each vertex, the reservoirs of its children are located below the reservoir of this vertex, and the vertex is connected with each of the children by a pipe through which water can flow downwards.

Mike wants to do the following operations with the tree:

Fill vertex v with water. Then v and all its children are filled with water.
Empty vertex v. Then v and all its ancestors are emptied.
Determine whether vertex v is filled with water at the moment.

Initially all vertices of the tree are empty.

Mike has already compiled a full list of operations that he wants to perform in order. Before experimenting with the tree Mike decided to run the list through a simulation. Help Mike determine what results will he get after performing all the operations.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 500000) — the number of vertices in the tree. Each of the following n - 1 lines contains two space-separated numbers a _i, b _i (1 ≤ a _i, b _i ≤ n, a _i ≠ b _i) — the edges of the tree.

The next line contains a number q (1 ≤ q ≤ 500000) — the number of operations to perform. Each of the following q lines contains two space-separated numbers c _i (1 ≤ c _i ≤ 3), v _i (1 ≤ v _i ≤ n), where c _i is the operation type (according to the numbering given in the statement), and v _i is the vertex on which the operation is performed.

It is guaranteed that the given graph is a tree.

Output

For each type 3 operation print 1 on a separate line if the vertex is full, and 0 if the vertex is empty. Print the answers to queries in the order in which the queries are given in the input.

【思路】树链剖分

  1 #include <algorithm>
  2 #include <cmath>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <list>
  6 #include <map>
  7 #include <iostream>
  8 #include <queue>
  9 #include <set>
 10 #include <stack>
 11 #include <string>
 12 #include <vector>
 13 #include <iomanip>
 14 #define LL long long
 15 #define inf 0x3f3f3f3f
 16 #define F first
 17 #define S second
 18 #define lson  rt << 1
 19 #define rson  rt << 1 | 1
 20 using namespace std;
 21 
 22 const int maxn = 5e5 + 7;
 23 int n, m, q;
 24 
 25 int head[maxn];
 26 struct node
 27 {
 28     int fr, v, w, nxt;
 29 }e[maxn * 2];
 30 
 31 int num = 0;
 32 void add(int t1, int t2)
 33 {
 34     e[++num].fr = t1;
 35     e[num].v = t2;
 36     e[num].nxt = head[t1];
 37     head[t1] = num;
 38 }
 39 
 40 int son[maxn], sz[maxn], f[maxn], dep[maxn];//重儿子， 子结点个数， 父节点， 节点深度
 41 //int dis[maxn];边权
 42 void dfs1(int u, int fa)
 43 {
 44     sz[u] = 1;
 45     for (int i = head[u]; i; i = e[i].nxt)
 46     {
 47         int v = e[i].v;
 48         if (v == fa) continue;
 49         dep[v] = dep[u] + 1;
 50         //dis[v] = dis[u] + e[i].w;
 51         f[v] = u;
 52         dfs1(v, u);
 53         sz[u] += sz[v];
 54         if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
 55     }
 56 }
 57 int top[maxn], id[maxn];
 58 int a[maxn], w[maxn]; //节点所在链的顶端， 节点dfs序编号， dfs序编号下当前点的点值， 节点的点值
 59 
 60 int cnt = 0;
 61 void dfs2(int u, int t)
 62 {
 63     id[u] = ++cnt;
 64     a[cnt] = w[u];
 65     top[u] = t;
 66     if (son[u]) dfs2(son[u], t);
 67     for (int i = head[u]; i; i = e[i].nxt)
 68     {
 69         int v = e[i].v;
 70         if (v == f[u] || v == son[u]) continue;
 71         dfs2(v, v);
 72     }
 73 }
 74 
 75 int tree[maxn * 4];
 76 int lz[maxn * 4];
 77 
 78 void push_down(int node, int l, int r) {
 79     if (lz[node] != -1) {
 80         lz[node * 2] = lz[node];
 81         lz[node * 2 + 1] = lz[node];
 82         tree[node * 2] =  tree[node];
 83         tree[node * 2 + 1] = tree[node];
 84         lz[node] = -1;
 85     }
 86     return;
 87 }
 88 
 89 void update_range(int node, int l, int r, int L, int R, int add) {
 90     if (l <= L && r >= R) {
 91         lz[node] = add;
 92         tree[node] = add;
 93         return;
 94     }
 95     push_down(node, L, R);
 96     int mid = (L + R) / 2;
 97     if (mid >= l) update_range(node * 2, l, r, L, mid, add);
 98     if (mid < r) update_range(node * 2 + 1, l, r, mid + 1, R, add);
 99 }
100 
101 int query(int node, int L, int R, int l, int r) {
102     if (L == R) return tree[node];
103     push_down(node, L, R);
104     int mid = (L + R) / 2;
105     if (mid >= l) return query(node * 2, L, mid, l, r) ;
106     else return query(node * 2 + 1, mid + 1, R, l, r) ;
107 }
108 void update_chain(int x, int y, int z)
109 {
110     int fx = top[x], fy = top[y];
111     while (fx != fy)
112     {
113         if (dep[fx] < dep[fy])
114         {
115             swap(x, y);
116             swap(fx, fy);
117         }
118         update_range(1, id[fx], id[x], 1, n, z);
119         x = f[fx], fx = top[x];
120     }
121     if (id[x] > id[y]) swap(x, y);
122     update_range(1, id[x], id[y], 1, n, z);
123 }
124 
125 int main()
126 {
127     ios::sync_with_stdio(false);
128     cin.tie(0);
129     cin >> n;
130     int ta, tb;
131     memset(lz, -1, sizeof(lz));
132     for (int i = 1; i <= n - 1; i++)
133     {
134         cin >> ta >> tb;
135         add(ta, tb);
136         add(tb, ta);
137     }
138     f[1] = 1;
139     dfs1(1, 0);
140     dfs2(1, 1);
141     cin >> q;
142     int c, x;
143     for (int i = 1; i <= q; i++)
144     {
145         cin >> c >> x;
146         switch (c)
147         {
148         case 1:
149         {
150             update_range(1, id[x], id[x] + sz[x] - 1, 1, n, 1);
151             break;
152         }
153         case 2:
154         {
155             update_chain(1, x, 0);
156             break;
157         }
158         case 3:
159         {
160             cout << query(1, 1, n, id[x], id[x]) << endl;
161         }
162         }
163     }
164 }