ZOJ 3686 A Simple Tree Problem

A Simple Tree Problem

---

Time Limit: 3 Seconds      Memory Limit: 65536 KB

---

Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0.

We define this kind of operation: given a subtree, negate all its labels.

And we want to query the numbers of 1's of a subtree.

Input

Multiple test cases.

First line, two integer N and M, denoting the numbers of nodes and numbers of operations and queries.(1<=N<=100000, 1<=M<=10000)

Then a line with N-1 integers, denoting the parent of node 2..N. Root is node 1.

Then M lines, each line are in the format "o node" or "q node", denoting we want to operate or query on the subtree with root of a certain node.

Output

For each query, output an integer in a line.

Output a blank line after each test case.

Sample Input

3 2
1 1
o 2
q 1

Sample Output

1

---

Author: CUI, Tianyi
Contest: ZOJ Monthly, March 2013

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 
 using namespace std;
 
 vector<int> g[];
 int n,m,id;
 
 const int maxn=;
 
 struct Interval
 {
     int from,to;
 }I[];
 
 void dfs(int node)
 {
     I[node].from=id;
     id++;
     int t=g[node].size();
     for(int i=;i<t;i++)
     {
         dfs(g[node][i]);
     }
     I[node].to=id-;
 }
 
 int m0[maxn<<],m1[maxn<<],xxo[maxn<<];
 
 void push_up(int rt)
 {
     m0[rt]=m0[rt<<]+m0[rt<<|];
     m1[rt]=m1[rt<<]+m1[rt<<|];
 }
 
 void push_down(int rt)
 {
     if(xxo[rt])
     {
         xxo[rt<<]^=; xxo[rt<<|]^=;
         swap(m0[rt<<|],m1[rt<<|]);swap(m0[rt<<],m1[rt<<]);
         xxo[rt]=;
     }
 }
 
 void build(int l,int r,int rt)
 {
     xxo[rt]=;m0[rt]=;m1[rt]=;
     if(l==r)
     {
         m0[rt]=; m1[rt]=;
         return ;
     }
     int m=(l+r)>>;
     push_down(rt);
     build(lson); build(rson);
     push_up(rt);
 }
 
 void update(int L,int R,int l,int r,int rt)
 {
     if(L<=l&&r<=R)
     {
         xxo[rt]^=;
         swap(m0[rt],m1[rt]);
         return ;
     }
     int m=(l+r)>>;
     push_down(rt);
     if(L<=m) update(L,R,lson);
     if(R>m) update(L,R,rson);
     push_up(rt);
 }
 
 int query(int L,int R,int l,int r,int rt)
 {
     if(L<=l&&r<=R)
     {
         push_down(rt);
         return m1[rt];
     }
     int m=(l+r)>>,ret=;
     push_down(rt);
     if(L<=m) ret+=query(L,R,lson);
     if(R>m) ret+=query(L,R,rson);
     push_up(rt);
     return ret;
 }
 
 int main()
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(int i=;i<=n+;i++) g[i].clear();
         memset(m0,,sizeof(m0));
         memset(m1,,sizeof(m1));
         memset(xxo,,sizeof(xxo));
         for(int i=;i<=n;i++)
         {
             int a;
             scanf("%d",&a);
             g[a].push_back(i);
         }
         id=;
         dfs();
         build(,n,);
         while(m--)
         {
             char cmd[]; int a;
             scanf("%s%d",cmd,&a);
             if(cmd[]=='o')   update(I[a].from,I[a].to,,n,);
             else if(cmd[]=='q')   printf("%d\n",query(I[a].from,I[a].to,,n,));
         }
         putchar();
     }
     return ;
 }