Codeforces Round #442 (Div. 2) E. Danil and a Part-time Job （dfs序树型转线性 线段树区间修改区间查询）

E. Danil and a Part-time Job

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Danil decided to earn some money, so he had found a part-time job. The interview have went well, so now he is a light switcher.

Danil works in a rooted tree (undirected connected acyclic graph) with n vertices, vertex 1 is the root of the tree. There is a room in each vertex, light can be switched on or off in each room. Danil's duties include switching light in all rooms of the subtree of the vertex. It means that if light is switched on in some room of the subtree, he should switch it off. Otherwise, he should switch it on.

Unfortunately (or fortunately), Danil is very lazy. He knows that his boss is not going to personally check the work. Instead, he will send Danil tasks using Workforces personal messages.

There are two types of tasks:

1. pow v describes a task to switch lights in the subtree of vertex v.
2. get v describes a task to count the number of rooms in the subtree of v, in which the light is turned on. Danil should send the answer to his boss using Workforces messages.

A subtree of vertex v is a set of vertices for which the shortest path from them to the root passes through v. In particular, the vertex v is in the subtree of v.

Danil is not going to perform his duties. He asks you to write a program, which answers the boss instead of him.

Input

The first line contains a single integer n (1 ≤ n ≤ 200 000) — the number of vertices in the tree.

The second line contains n - 1 space-separated integers p2, p3, ..., pn (1 ≤ pi < i), where pi is the ancestor of vertex i.

The third line contains n space-separated integers t1, t2, ..., tn (0 ≤ ti ≤ 1), where ti is 1, if the light is turned on in vertex i and 0 otherwise.

The fourth line contains a single integer q (1 ≤ q ≤ 200 000) — the number of tasks.

The next q lines are get v or pow v (1 ≤ v ≤ n) — the tasks described above.

Output

For each task get v print the number of rooms in the subtree of v, in which the light is turned on.

Example

input

41 1 11 0 0 19get 1get 2get 3get 4pow 1get 1get 2get 3get 4

output

20012110

Note

The tree before the task pow 1.

The tree after the task pow 1.

#include <bits/stdc++.h>using namespace std;#define maxn 200005int c[maxn * 4], lazy[maxn * 4], lft[maxn], rght[maxn], tot, sz[maxn];vector<vector<int> > g(maxn);void pushdown(int o, int l, int r){if(l == r || lazy[o] == 0) return;int mid = l + r >> 1;c[o << 1] = mid - l + 1 - c[o << 1];c[o << 1 | 1] = r - mid - c[o << 1 | 1];lazy[o << 1] ^= 1;lazy[o << 1 | 1] ^= 1;lazy[o] = 0;}void add(int o, int l, int r, int L, int R){pushdown(o, l, r);if(l >= L && r <= R){c[o] = r - l + 1 - c[o];lazy[o] ^= 1;return ;}int mid = l + r >> 1;if(mid >= L) add(o << 1, l, mid, L, R);if(mid < R) add(o << 1 | 1, mid + 1, r, L, R);c[o] = c[o << 1] + c[o << 1 | 1];}void dfs(int x){lft[x] = tot;for(int i = 0; i < g[x].size(); ++i){tot++;dfs(g[x][i]);}rght[x] = tot;}int getsum(int o, int l, int r, int L, int R){pushdown(o, l, r);if(l >= L && r <= R){return c[o];}int mid = l + r >> 1;int ans = 0;if(mid >= L) ans += getsum(o << 1, l, mid, L, R);if(mid < R) ans += getsum(o << 1 | 1, mid + 1, r, L, R);return ans;}int main(){int n, q, u;cin>>n;for(int i = 2; i <= n; ++i){scanf("%d", &u);g[u].push_back(i);}memset(c, 0, sizeof(c));memset(lazy, 0, sizeof(lazy));for(int i = 1; i <= n; ++i){scanf("%d", &sz[i]);}tot = 1;dfs(1);for(int i = 1; i <= n; ++i){if(sz[i]) add(1, 1, n, lft[i], lft[i]);}char s[5];scanf("%d", &q);while(q--){scanf("%s", s);if(s[0] == 'p'){scanf("%d", &u);add(1, 1, n, lft[u], rght[u]);sz[u] ^= 1;}else{scanf("%d", &u);printf("%d\n", getsum(1, 1, n, lft[u], rght[u]));}}}/*题意：一棵树，2e5个节点，每个节点上一个灯泡，2e5次操作，每次操作要么将以某个节点为子树上的所有灯泡全部翻转（开变关，关变开），要么询问以某个节点为子树上有多少个灯泡是亮的。思路：裸的dfs树形转线性+线段树区间修改和查询。。。。这题搞个dfs序就行了，维护一下子树上最小编号和最大编号，然后对应到线性区间上。区间修改和查询应该是线段树的基本操作了，用lazy记下翻转标记，*/