A Simple Tree Problem
This problem will be judged on ZJU. Original ID: 3686
64-bit integer IO format: %lld Java class name: Main
None
Graph Theory
2-SAT
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Math
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Tag it!
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
题目大意：有棵根节点为1的树，共有n个节点。有m次询问。第二行为从2--->n各个节点对应的父亲节点编号。下面的m行是询问，o ai表示将节点为ai的子树所有节点的值进行异或即0变1,1变0。q ai表示询问目前该子树的节点的和值为多少。
解题思路：其实这个题目重点在如何将多子树转化成线段树进行操作。我们如果重新将树编号，那么可以让每个节点对应一段区间。从根节点1开始深搜，编号为1，每当搜到一个节点，就让编号的值加1，让这个编号等于该节点的区间左端点，等把该节点的所有子节点访问完后，将这时的编号赋值给该节点的区间右端点。这时这个区间内的所有节点都是该节点的子节点。 后边就是区间更新的问题了。
#include<bits/stdc++.h>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=1e5+50;
vector<int>G[maxn];
int Lt[maxn],Rt[maxn];
int sumv[maxn*4],lazy[maxn*4];
int n,cn;
void dfs(int u){
Lt[u]=++cn;
int v;
for(int i=0;i<G[u].size();i++){
v=G[u][i];
dfs(v);
}
Rt[u]=cn;
}
void build(int rt,int L,int R){
sumv[rt]=lazy[rt]=0;
if(L==R)
return ;
build(lson);
build(rson);
}
void PushDown(int rt,int L,int R){
if(lazy[rt]){
lazy[rt*2]^=1;
lazy[rt*2+1]^=1;
sumv[rt*2]=(mid-L+1)-sumv[rt*2];
sumv[rt*2+1]=(R-mid)-sumv[rt*2+1];
lazy[rt]=0;
}
}
void PushUp(int rt){
sumv[rt]=sumv[rt*2]+sumv[rt*2+1];
}
void update(int rt,int L,int R,int l_ran,int r_ran){
if(l_ran<=L&&R<=r_ran){
lazy[rt]^=1;
sumv[rt]=R-L+1-sumv[rt];
return ;
}
PushDown(rt,L,R);
if(l_ran<=mid){
update(lson,l_ran,r_ran);
}
if(r_ran>mid){
update(rson,l_ran,r_ran);
}
PushUp(rt);
}
int query(int rt,int L,int R,int l_ran,int r_ran){
if(l_ran<=L&&R<=r_ran){
return sumv[rt];
}
int ret=0;
PushDown(rt,L,R); //lazy下放
if(l_ran<=mid){
ret+=query(lson,l_ran,r_ran);
}
if(r_ran>mid){
ret+=query(rson,l_ran,r_ran);
}
return ret;
}
int main(){
char s[4];
int m,subt,a,res;
while(scanf("%d%d",&n,&m)!=EOF){
build(1,1,n);
for(int i=0;i<=n;i++)
G[i].clear();
for(int i=2;i<=n;i++){
scanf("%d",&a);
G[a].push_back(i);
}
cn=0;
dfs(1);
for(int i=0;i<m;i++){
scanf("%s%d",s,&subt);
if(s[0]=='o'){
update(1,1,n,Lt[subt],Rt[subt]);
}else{
res=query(1,1,n,Lt[subt],Rt[subt]);
printf("%d\n",res);
}
}printf("\n"); }
return 0;
} /*
6 5
1 2 1 4 4
o 4
q 4
q 5
q 6
q 1
*/