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 = 100010;
int n,m,id;
vector<int>g[maxn];
int len[maxn<<2],cnt[maxn<<2],mark[maxn<<2];
struct node{
    int lt,rt;
}tree[maxn<<2];
void dfs(int u){
    tree[u].lt = ++id;
    for(int i = 0; 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] = 0;
    len[v] = rt-lt+1;
    if(lt == rt) return;
    int mid = (lt+rt)>>1;
    build(lt,mid,v<<1);
    build(mid+1,rt,v<<1|1);
}
void push_down(int v){
    if(mark[v]){
        cnt[v<<1] = len[v<<1]-cnt[v<<1];
        mark[v<<1] ^= 1;
        cnt[v<<1|1] = len[v<<1|1]-cnt[v<<1|1];
        mark[v<<1|1] ^= 1;
        mark[v] = 0;
    }
}
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] ^= 1;
        return;
    }
    push_down(v);
    int mid = (lt+rt)>>1;
    if(x <= mid) update(x,y,lt,mid,v<<1);
    if(y > mid) update(x,y,mid+1,rt,v<<1|1);
    cnt[v] = cnt[v<<1]+cnt[v<<1|1];
}
int query(int x,int y,int lt,int rt,int v){
    int ans = 0;
    if(x <= lt && rt <= y){
        return cnt[v];
    }
    push_down(v);
    int mid = (lt+rt)>>1;
    if(x <= mid) ans += query(x,y,lt,mid,v<<1);
    if(y > mid) ans += query(x,y,mid+1,rt,v<<1|1);
    return ans;
}
int main(){
    int i,temp;
    char s;
    while(~scanf("%d%d",&n,&m)){
        for(i = 0; i <= n; i++)
            g[i].clear();
        for(i = 2; i <= n; i++){
            scanf("%d",&temp);
            g[temp].push_back(i);
        }
        id = 0;
        dfs(1);
        build(1,n,1);
        while(m--){
            cin>>s>>temp;
            if(s == 'o'){
                update(tree[temp].lt,tree[temp].rt,1,n,1);
            }else cout<<query(tree[temp].lt,tree[temp].rt,1,n,1)<<endl;
        }
        cout<<endl;
    }
    return 0;
}