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.
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 0otherwise.
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.
For each task get v print the number of rooms in the subtree of v, in which the light is turned on.
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
2
0
0
1
2
1
1
0
题意：n点 输入每个点的值0或1 输入每个点的父节点 m次操作 pow操作改变q节点后面的全部节点包括q节点 get操作是问q节点的子节点包括q一共有多少个1
这个题要用线段树+dfs,意思就是为每个点分配一个区间，区间大的包含区间小的
这个题要注意的是
每个点可能变换多次，你不需要全都变换，查询时候在一次性的变完就行，否则可能会超时，反正我链表这样做超时了，数组我好像runtime error了 最后也懒得改了，又敲了一遍过的，然后就没什么注意的了，这题就是这样
丑陋的代码
#include <iostream>
#include <vector>
using namespace std;
void pushdown(int i);
const int N = 2e5 + 10;
int a[N],in[N],out[N],pre[N],cnt = 0;
vector<int> p[N];
struct node
{
int l = 0,r = 0;
int sum = 0;
int lazy = 0;
}pp[N<<2];
void dfs(int n)
{
for(int i = 0; i < p[n].size(); ++i)
{
int d = p[n][i];
cnt ++;
in[d] = cnt;
pre[cnt] = d;
dfs(d);
out[d] = cnt;
}
}
void build(int i)
{
int mid = (pp[i].l+pp[i].r)>>1;
int l = pp[i].l;
int r = pp[i].r;
if(pp[i].l == pp[i].r)
{
pp[i].sum = a[pre[l]];
return ;
}
pp[(i<<1)].l = l;
pp[(i<<1)].r = mid;
pp[(i<<1)+1].l = mid+1;
pp[(i<<1)+1].r = r;
build((i<<1));
build((i<<1)+1);
pp[i].sum += pp[i<<1].sum + pp[(i<<1)+1].sum;
}
void update(int l,int r,int i)
{
int mid = (pp[i].l+pp[i].r) >> 1;
if(pp[i].l == l && pp[i].r == r)
{
pp[i].sum = pp[i].r - pp[i].l + 1 - pp[i].sum;
pp[i].lazy ^= 1;
pushdown(i);
//这个地方要注意pushdowm的操作位置
return ;
}
if(r <= mid)
{
update(l,r, i<<1);
}
else if(l >= mid + 1)
{
update(l,r,(i<<1)+1);
}
else
{
update(l, mid, i<<1);
update(mid+1, r, (i<<1)+1);
}
pp[i].sum = pp[i<<1].sum + pp[(i<<1)+1].sum;
}
void pushdown(int i)
{
if(pp[i].lazy)
{
pp[i].sum = pp[i].r - pp[i].l + 1 - pp[i].sum;
pp[i<<1].lazy ^= 1;
pp[(i<<1)+1].lazy ^= 1;
pp[i].lazy ^= 1;
}
}
int query(int l,int r,int i)
{
int mid = (pp[i].l + pp[i].r)>>1;
pushdown(i);
if(pp[i].l == l && pp[i].r == r)
return pp[i].sum;
if(r <= mid)
{
return query(l, r, i<<1);
}
else if(l >= mid + 1)
{
return query(l, r, (i<<1)+1);
}
else
{
return query(l, mid, i<<1) + query(mid+1, r, (i<<1)+1);
}
}
int main()
{
int n,i,j,d,x,ppp;
char str[20];
scanf("%d",&n);
for(i = 2; i <= n; ++i)
{
scanf("%d",&d);
p[d].push_back(i);
}
for(i = 1; i <= n; ++i)
scanf("%d",&a[i]);
pre[1] = 1;
cnt = 1;
in[1] = 1;
out[1] = n;
pp[1].l = 1;
pp[1].r = n;
dfs(1);
build(1);
scanf("%d",&ppp);
while( ppp-- )
{
scanf("%s",str);
if(str[0] == 'g')
{
scanf("%d",&x);
printf("%d ",query(in[x], out[x], 1));
}
else
{
scanf("%d",&x);
update(in[x], out[x], 1);
}
}
return 0;
}
注意 update的时候假如(pp[i].l == l && pp[i].r == r)这个时候一定要更新当前节点 后面的节点可以不更新，但当前节点必须要更新
具体原因数据在这里 你画一下图就明白了
10 1 2 3 3 5 5 7 7 8 0 0 0 0 1 1 1 1 0 0 10 pow 3 get 1 pow 9 get 1 get 1 get 8 pow 8 pow 4 get 10 pow 2