xtu数据结构 I. A Simple Tree Problem

I. A Simple Tree Problem

Time Limit: 3000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

 

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

 解题：利用dfs记录时间戳，也就是记录区间长度，恰好的包含关系，为建立线段树带来很大的方便，不需要建立N棵线段树，但以某一点为根节点的字孩子区间一定是父节点的区间的子区间。这样好的性质，恰好可以映射到线段树上。。。。。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <vector>
 #include <climits>
 #include <algorithm>
 #include <cmath>
 #define LL long long
 #define INF 0x3f3f3f
 using namespace std;
 const int maxn = ;
 int n,m,id;
 vector<int>g[maxn];
 int len[maxn<<],cnt[maxn<<],mark[maxn<<];
 struct node{
     int lt,rt;
 }tree[maxn<<];
 void dfs(int u){
     tree[u].lt = ++id;
     for(int i = ; i < g[u].size(); i++){
         dfs(g[u][i]);
     }
     tree[u].rt = id;
 }
 void build(int lt,int rt,int v){
     cnt[v] = mark[v] = ;
     len[v] = rt-lt+;
     if(lt == rt) return;
     int mid = (lt+rt)>>;
     build(lt,mid,v<<);
     build(mid+,rt,v<<|);
 }
 void push_down(int v){
     if(mark[v]){
         cnt[v<<] = len[v<<]-cnt[v<<];
         mark[v<<] ^= ;
         cnt[v<<|] = len[v<<|]-cnt[v<<|];
         mark[v<<|] ^= ;
         mark[v] = ;
     }
 }
 void update(int x,int y,int lt,int rt,int v){
     if(lt >= x && rt <= y){
         cnt[v] = len[v] - cnt[v];
         if(lt == rt) return;
         mark[v] ^= ;
         return;
     }
     push_down(v);
     int mid = (lt+rt)>>;
     if(x <= mid) update(x,y,lt,mid,v<<);
     if(y > mid) update(x,y,mid+,rt,v<<|);
     cnt[v] = cnt[v<<]+cnt[v<<|];
 }
 int query(int x,int y,int lt,int rt,int v){
     int ans = ;
     if(x <= lt && rt <= y){
         return cnt[v];
     }
     push_down(v);
     int mid = (lt+rt)>>;
     if(x <= mid) ans += query(x,y,lt,mid,v<<);
     if(y > mid) ans += query(x,y,mid+,rt,v<<|);
     return ans;
 }
 int main(){
     int i,temp;
     char s;
     while(~scanf("%d%d",&n,&m)){
         for(i = ; i <= n; i++)
             g[i].clear();
         for(i = ; i <= n; i++){
             scanf("%d",&temp);
             g[temp].push_back(i);
         }
         id = ;
         dfs();
         build(,n,);
         while(m--){
             cin>>s>>temp;
             if(s == 'o'){
                 update(tree[temp].lt,tree[temp].rt,,n,);
             }else cout<<query(tree[temp].lt,tree[temp].rt,,n,)<<endl;
         }
         cout<<endl;
     }
     return ;
 }