CodeForces 877E Danil and a Part-time Job(dfs序+线段树)

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
inputCopy
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
outputCopy
2
0
0
1
2
1
1
0
Note

The tree before the task pow 1.

The tree after the task pow 1.

题意：给出一棵树初始的节点值，给出两种操作，一种为子树异或1，另一种为统计子树中一的个数

题解：  可以用dfs序+线段树解决，异或标记可以通过tag[x]^=1来进行传递，每次更改相当于将区间内所有的1改为0，所有的0改为1，这样相当于把sum改为size-sum，可以O(1)实现

代码如下：

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lson root<<1
#define rson root<<1|1
using namespace std;

struct node
{
    int l,r,lazy,sum;
} tr[];

vector<int> g[];
int id[],size[],c[],w[],tot;

void push_up(int root)
{
    tr[root].sum=tr[lson].sum+tr[rson].sum;
}

void push_down(int root)
{
    int mid=(tr[root].l+tr[root].r)>>;
    tr[lson].sum=(mid-tr[root].l+)-tr[lson].sum;
    tr[lson].lazy=^tr[lson].lazy;
    tr[rson].sum=(tr[root].r-mid)-tr[rson].sum;
    tr[rson].lazy=^tr[rson].lazy;
    tr[root].lazy=;
}

void build(int root,int l,int r)
{
    if(l==r)
    {
        tr[root].l=l;
        tr[root].r=r;
        tr[root].sum=w[l];
        return ;
    }
    tr[root].l=l;
    tr[root].r=r;
    int mid=(l+r)>>;
    build(lson,l,mid);
    build(rson,mid+,r);
    push_up(root);
}

void update(int root,int l,int r,int val)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        tr[root].sum=(tr[root].r-tr[root].l+)-tr[root].sum;
        tr[root].lazy=tr[root].lazy^;
        return ;
    }
    if(tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>;
    if(l>mid)
    {
        update(rson,l,r,val);
    }
    else
    {
        if(mid>=r)
        {
            update(lson,l,r,val);
        }
        else
        {
            update(lson,l,mid,val);
            update(rson,mid+,r,val);
        }
    }
    push_up(root);
}

int query(int root,int l,int r)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        return tr[root].sum;
    }
    if(tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>;
    if(mid<l)
    {
        return query(rson,l,r);
    }
    else
    {
        if(mid>=r)
        {
            return query(lson,l,r);
        }
        else
        {
            return query(lson,l,mid)+query(rson,mid+,r);
        }
    }
}

void dfs(int now,int f)
{
    id[now]=++tot;
    w[tot]=c[now];
    size[now]=;
    for(int i=; i<g[now].size(); i++)
    {
        if(g[now][i]==f)
        {
            continue;
        }
        dfs(g[now][i],now);
        size[now]+=size[g[now][i]];
    }
}

void sub_update(int u,int val)
{
    update(,id[u],id[u]+size[u]-,val);
}

int sub_query(int u)
{
    return query(,id[u],id[u]+size[u]-);
}

int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=; i<=n; i++)
    {
        int to;
        scanf("%d",&to);
        g[to].push_back(i);
        g[i].push_back(to);
    }
    for(int i=; i<=n; i++)
    {
        scanf("%d",&c[i]);
    }
    dfs(,);
    build(,,n);
    scanf("%d",&m);
    char s[];
    int val;
    for(int i=; i<=m; i++)
    {
        scanf("\n%s %d",s,&val);
        if(s[]=='g')
        {
            printf("%d\n",sub_query(val));
        }
        else
        {
            sub_update(val,);
        }
    }
}