Codeforces Round #442 (Div. 2) E. Danil and a Part-time Job DFS序+树链剖分+线段树区间^

传送门

题意：

给你一棵根为1的树, 每个点的权值是0或1, get x表示求x子树的1的个数, pow x表示把x的子树1变0, 0变1 

做法：很容易看出来这是树链剖分，再带上线段树的区间异或就好了- -

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scanf("%I64d",&x)
#define S_2(x,y) scanf("%I64d%I64d",&x,&y)
#define S_3(x,y,z) scanf("%I64d%I64d%I64d",&x,&y,&z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=1e19+10;
//const int dx[]={-1,0,1,0,1,-1,-1,1};
//const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=2e3+10;
const int maxx=1e6+10;
const double EPS=1e-10;
const double eps=1e-10;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
/*
void readString(string &s)
{
	static char str[maxx];
	scanf("%s", str);
	s = str;
}
*/

struct node
{
    int tl,tr,lazy,val;
    node(int x=0,int y=0,int z=0):
        lazy(0),tl(x),tr(y),val(z) {}
} tree[800020];

int a[200005];

void pushup(int id)
{
    tree[id].val=tree[id<<1].val+tree[id<<1|1].val;
}

void build(int id,int tl,int tr)
{
    tree[id]=node(tl,tr,0);
    if(tl==tr)tree[id].val=a[tl];
    else
    {
        int tm=(tr+tl)>>1;
        build(id<<1,tl,tm);
        build(id<<1|1,tm+1,tr);
        pushup(id);
    }
}

void pushdown(int id)
{
    if(!tree[id].lazy)return;
    tree[id<<1].lazy^=1,tree[id<<1|1].lazy^=1;
    tree[id<<1].val=tree[id<<1].tr-tree[id<<1].tl+1-tree[id<<1].val;
    tree[id<<1|1].val=tree[id<<1|1].tr-tree[id<<1|1].tl+1-tree[id<<1|1].val;
    tree[id].lazy=0;
}


void update(int id,int ql,int qr)
{
    int tl=tree[id].tl,tr=tree[id].tr;
    if(ql>tr || qr<tl)return;
    if(ql<=tl && tr<=qr)
    {
        tree[id].lazy^=1;
        tree[id].val=tr-tl+1-tree[id].val;
        return;
    }
    pushdown(id);
    update(id<<1,ql,qr);
    update(id<<1|1,ql,qr);
    pushup(id);
}

int get_sum(int id,int ql,int qr)
{
    int tl=tree[id].tl,tr=tree[id].tr,ans=0;
    if(ql>tr || qr<tl)return 0;
    if(ql<=tl && tr<=qr)return tree[id].val;
    pushdown(id);
    ans+=get_sum(id<<1,ql,qr);
    ans+=get_sum(id<<1|1,ql,qr);
    pushup(id);
    return ans;
}
/// dfs序+树链剖分
int pre[200005],siz[200005],dep[200005];
int top[200005],node_id[200005],heavy[200005];
int lson[200005],rson[200005],idx=0;

struct
{
    int nex,nex_node;
} edge[200005];

int head[200005],cont=0;

void add_edge(int now,int nex)
{
    edge[cont].nex=nex;
    edge[cont].nex_node=head[now];
    head[now]=cont++;
}

int dfs(int x,int fa,int depth)
{
    dep[x]=depth,pre[x]=fa;
    int res=1,maxx=0;
    for(int i=head[x]; ~i; i=edge[i].nex_node)
    {
        int nex=edge[i].nex;
        if(nex==fa)continue;
        int t=dfs(nex,x,depth+1);
        if(t>maxx)maxx=t,heavy[x]=nex;
        res+=t;
    }
    return siz[x]=res;
}

void slpf(int x,int fa,int tp)
{
    lson[x]=node_id[x]=++idx,top[x]=tp;
    if(heavy[x])slpf(heavy[x],x,tp);
    for(int i=head[x]; ~i; i=edge[i].nex_node)
    {
        int nex=edge[i].nex;
        if(nex==fa)continue;
        if(nex!=heavy[x])slpf(nex,x,nex);
    }
    rson[x]=idx;
}

int b[200005];
int n;
void solve()
{
    W(s_1(n)!=EOF)
    {
        me(head,-1);
        FOR(2,n,i)
        {
            int x;
            s_1(x);
            add_edge(x,i);
        }
        FOR(1,n,i)
            s_1(b[i]);
        dfs(1,0,1);
        slpf(1,0,1);
        FOR(1,n,i)
            a[lson[i]]=b[i];
        build(1,1,n);
        int q;
        s_1(q);
        W(q--)
        {
            int x;
            char op[10];
            scanf("%s%d",op,&x);
            if(op[0]=='g')
                printf("%d\n",get_sum(1,lson[x],rson[x]));
            else update(1,lson[x],rson[x]);
        }
    }
}

int main()
{
    //freopen( "in.txt" , "r" , stdin );
    //freopen( "data.txt" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++)
    {
        //printf("Case #%d: ",cas);
        solve();
    }
}