高级数据结构第六章E . 苹果树 (dfs+树状数组)
思路:
经典套路,通过dfs序将树上修改转化为线性修改,这样问题就转化为了单点修改,区间查询,用树状数组维护。
类似题
代码:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
ll res = 1;
while(b)
{
if(b & 1)res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=1e5+100;
struct BIT{
ll n,tr[maxn];
void init(){
memset(tr,0,sizeof tr);
}
ll lowbit(ll x){
return x&-x;
}
void update(ll pos,ll val){
while(pos<=n){
tr[pos]+=val;pos+=lowbit(pos);
}
}
ll qask(ll pos){
ll res=0;
while(pos){
res+=tr[pos];pos-=lowbit(pos);
}
return res;
}
};
int n,m,a[maxn];
vector<int>g[maxn];
ll in[maxn],out[maxn],timetmp;
void dfs(int u,int fa){//dfs记录dfs序列
in[u]=++timetmp;
for(int i=0;i<g[u].size();i++){
int j=g[u][i];
if(j==fa) continue;
dfs(j,u);
}
out[u]=timetmp;
}
int main()
{
n=read;
rep(i,1,n-1){
int u=read,v=read;
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1,-1);
BIT t;
t.n=1e5;t.init();
rep(i,1,n) t.update(in[i],1),a[i]=1;
m=read;
rep(i,1,m){
char op[2];int x;
cin>>op;x=read;
if(op[0]=='C'){
if(a[x]) t.update(in[x],-1);
else t.update(in[x],1);
a[x]=!a[x];
}
else{
printf("%lld\n",t.qask(out[x])-t.qask(in[x]-1));
}
}
return 0;
}
/*
**/