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:
- pow v describes a task to switch lights in the subtree of vertex v.
- 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
Copy
4 1 1 1 1 0 0 1 9 get 1 get 2 get 3 get 4 pow 1 get 1 get 2 get 3 get 4
output
Copy
2 0 0 1 2 1 1 0
题目大意 :
一颗点权只可能为0或1的树, 现在有两种操作, 第一个是将某个点的子树包括他本身的所有点的点权变成另一个值, 例如0 -> 1, 1 -> 0, 第二个操作是输出某个点的子树包括他本身的点权和
思路 :
树上子树的区间问题自然是用dfs序,维护的是区间和, 不难想到,一个区间进行操作的时候, 该区间的点权值和会变为该区间的长度 - 当前的和, 所以只要在更新的时候加一个标记,并且更新为偶数次的时候相当于不更新,通过将异或就可以实现上面的操作了
Accepted code
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define sc scanf
#define ls rt << 1
#define rs ls | 1
#define Min(x, y) x = min(x, y)
#define Max(x, y) x = max(x, y)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define pir pair <int, int>
#define MEM(x, b) memset(x, b, sizeof(x))
#define lowbit(x) ((x) & -(x))
#define P2(x) ((x) * (x))
typedef long long ll;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 100;
const int INF = 0x3f3f3f3f;
inline ll fpow(ll a, ll b){ ll r = 1, t = a; while (b){ if (b & 1)r = (r*t) % MOD; b >>= 1; t = (t*t) % MOD; }return r; }
struct Tree
{
int l, r, ans;
int len, lzy;
}t[MAXN * 4];
vector <int> G[MAXN << 1];
int n, in[MAXN], out[MAXN], ev[MAXN], vb[MAXN];
int val[MAXN], tot, q;
void Build(int rt, int l, int r) {
t[rt].l = l, t[rt].r = r;
t[rt].len = r - l + 1;
if (l == r) {
t[rt].ans = val[vb[l]]; // 记的是更新前的点权
return;
}
int mid = (l + r) >> 1;
Build(ls, l, mid);
Build(rs, mid + 1, r);
t[rt].ans = t[ls].ans + t[rs].ans;
}
void dfs(int x, int fa) {
ev[x] = ++tot, in[ev[x]] = tot; // dfs序
vb[ev[x]] = x;
for (auto it : G[x]) {
if (it != fa)
dfs(it, x);
}
out[ev[x]] = tot;
}
void Pushdown(int rt) {
t[ls].lzy ^= t[rt].lzy, t[rs].lzy ^= t[rt].lzy;
t[ls].ans = t[ls].len - t[ls].ans;
t[rs].ans = t[rs].len - t[rs].ans;
t[rt].lzy = 0;
}
void Update(int rt, int l, int r) {
if (t[rt].l >= l && t[rt].r <= r) {
t[rt].ans = t[rt].len - t[rt].ans; // 区间长度 - 当前和
t[rt].lzy ^= 1; // 奇数次更新,偶数次不更新,异或即可
return;
}
if (t[rt].lzy)
Pushdown(rt);
int mid = (t[rt].l + t[rt].r) >> 1;
if (mid >= r)
Update(ls, l, r);
else if (mid < l)
Update(rs, l, r);
else {
Update(ls, l, mid);
Update(rs, mid + 1, r);
}
t[rt].ans = t[ls].ans + t[rs].ans;
}
int Query(int rt, int l, int r) {
if (t[rt].l >= l && t[rt].r <= r)
return t[rt].ans;
if (t[rt].lzy)
Pushdown(rt);
int mid = (t[rt].l + t[rt].r) >> 1;
int ui = 0, vi = 0;
if (mid >= l)
ui = Query(ls, l, r);
if (mid < r)
vi = Query(rs, l, r);
return ui + vi;
}
int main()
{
cin >> n;
for (int i = 2; i <= n; i++) {
int tmp; sc("%d", &tmp);
G[i].push_back(tmp);
G[tmp].push_back(i);
}
for (int i = 1; i <= n; i++)
sc("%d", &val[i]);
dfs(1, 0);
Build(1, 1, n);
cin >> q;
while (q--) {
string op;
int tmp;
cin >> op >> tmp;
if (op[0] == 'p')
Update(1, in[ev[tmp]], out[ev[tmp]]); // 该点的子树区间
else
printf("%d\n", Query(1, in[ev[tmp]], out[ev[tmp]]));
}
return 0; // 改数组大小!!!
}