题目大意
给你一棵树， 求一点到根的路径上有多少个未标记点并全标记， 和询问一个点的子树内有多少已标记点和撤销标记
解题方法
1: install 操作
这个操作是求一点到根的路径上有多少个未标记点并全标记， 这种操作可以用树链剖分来解决，将已标记的点的权值设为1， 求和即可
2.uninstall 操作
这个操作询问一个点的子树内有多少已标记点和撤销标记， 同理上面的方法就好了。
代码实现
#include <bits/stdc++.h>
using namespace std;
template<class T>
inline void read(T &a){
T s = 0, w = 1;
char c = getchar();
while(c < '0' || c > '9') {if(c == '-') w = -1; c =getchar();}
while(c >= '0' && c <= '9') {s = (s << 1) + (s << 3) + (c ^ 48); c = getchar();}
a = s * w;
}
#define maxn 200100
#define maxm 200100
static int n, m;
static int net[maxm], to[maxm], head[maxm], tot;
inline void add(int x, int y){
net[++tot] = head[x], head[x] = tot, to[tot] = y;
}
/*-----------------------------------------------------------*/
static int fat[maxn], size[maxn], deep[maxn];
static int son[maxn];
void dfs1(int x, int fa){
fat[x] = fa;
size[x] = 1;
son[x] = 0;
deep[x] = deep[fa] + 1;
for (int i = head[x];i;i = net[i]){
int v = to[i];
if(v == fa) continue;
dfs1(v, x);
size[x] += size[v];
if(size[v] > size[son[x]]) son[x] = v;
}
}
static int tid[maxn], hhd[maxn], cnt, top[maxn];
void dfs2(int x, int t, int fa){
tid[x] = ++cnt; hhd[cnt] = x; top[x] = t;
if(!son[x]) return;
top[x] = t;
dfs2(son[x], t, x);
for (int i = head[x];i;i = net[i]){
int v = to[i];
if(v == fa || v == son[x]) continue;
dfs2(v, v, x);
}
}
/*-----------------------------------------*/
static int Sum[maxn * 8], lazy[maxn * 8];
#define ls rt << 1
#define rs rt << 1 | 1
inline void push_down(int rt, int l, int r){
int mid = (l + r) / 2;
Sum[ls] = (mid - l + 1) * lazy[rt];
Sum[rs] = (r - (mid + 1) + 1) * lazy[rt];
lazy[ls] = lazy[rt]; lazy[rs] = lazy[rt];
lazy[rt] = -1;
}
int ans = 0;
int Query(int rt, int l, int r, int L, int R){
if(L <= l && r <= R){
return Sum[rt];
}
else{
int mid = (l + r) / 2;
if(lazy[rt] != -1) push_down(rt, l, r);
int ans = 0;
if(L <= mid) ans += Query(ls, l, mid, L, R);
if(R > mid) ans += Query(rs, mid + 1, r, L, R);
Sum[rt] = Sum[ls] + Sum[rs];
return ans;
}
}
void Change(int rt, int l, int r, int L, int R, int d){
if(L <= l && r <= R){
Sum[rt] = (r - l + 1) * d;
lazy[rt] = d;
}
else{
int mid = (l + r) / 2;
if(lazy[rt] != -1) push_down(rt, l, r);
if(L <= mid) Change(ls, l , mid, L, R, d);
if(R > mid) Change(rs, mid + 1, r, L, R, d);
Sum[rt] = Sum[ls] + Sum[rs];
}
}
inline int Query_path(int x, int y){
int fx = top[x], fy = top[y];
int ans = 0;
while(fx != fy){
if(deep[x] < deep[y]) swap(x, y), swap(fx, fy);
ans += Query(1, 1, n, tid[fx], tid[x]);
x = fat[fx];
fx = top[x];
}
if(tid[x] < tid[y]) swap(x, y), swap(fx, fy);
ans += Query(1, 1, n, tid[y], tid[x]);
return ans;
}
inline void Change_path(int x, int y, int d){
int fx = top[x], fy = top[y];
int ans = 0;
while(fx != fy){
if(deep[x] < deep[y]) swap(x, y), swap(fx, fy);
Change(1, 1, n, tid[fx], tid[x], d);
x = fat[fx];
fx = top[x];
}
if(tid[x] < tid[y]) swap(x, y);
Change(1, 1, n, tid[y], tid[x], d);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("p2146.in","r", stdin);
freopen("p2146.out","w", stdout);
#endif
memset(lazy, -1, sizeof(lazy));
read(n);
for (int i = 1; i <= n-1; i++){
int x;
read(x);
add(x + 1, i + 1); add(i + 1, x + 1);
}
read(m);
dfs1(1, 0);
dfs2(1,1,0);
// build(1, 1, n);
for (int i = 1; i <= m; i++){
string s;
cin>>s;
// cout<<s<<endl;
if(s == "install"){
int x;
read(x); x++;
// printf("ss %d\n",i);
int ans = Query_path(1, x);
printf("%d\n", deep[x] - ans);
Change_path(1, x, 1);
}
else{
int x = 0;
read(x); x++;
int ans = Query(1, 1, n, tid[x], tid[x] + size[x] - 1);
printf("%d\n", ans);
Change(1, 1, n, tid[x], tid[x] + size[x] - 1, 0);
}
}
return 0;
}