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_backtypedef 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();    }}