题面
解析
上课讲稿上的例题
这道题是套路题，是括号序的应用，进入节点时打上$+1$标记, 退出时打上$-1$标记，这个是作为点权的系数
先看操作2， 需要更改父节点，就是把一段区间提取出来，插入另一个地方，显然可以用Splay维护，先提取区间，再把新父亲的$+1$点旋转至根，把区间挂在根的后继的左儿子上，再把这个节点旋转至根，以更新信息
对于操作1，求点到根的路径和，就是求括号序列的前缀和，该点对应的$+1$点或$-1$点的前缀和都可，我是把$-1$的点旋转至根，答案就是根的左儿子的子树和，因此需要维护子树和
最后还有操作3，显然打标记，问题在对于子树和的改变， 记该子树中$+1$标记有$cnt1$个，$-1$标记有$cnt_1$个，则子树和需要加$(cnt1 - cnt_1) * delta$，因此需要维护子树中$+1$与$-1$的个数
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 100004;
template<class T> void read(T &re)
{
re=0;
T sign=1;
char tmp;
while((tmp=getchar())&&(tmp<'0'||tmp>'9')) if(tmp=='-') sign=-1;
re=tmp-'0';
while((tmp=getchar())&&(tmp>='0'&&tmp<='9')) re=(re<<3)+(re<<1)+(tmp-'0');
re*=sign;
}
int n, m, root, cnt, a[maxn], fir[maxn], sec[maxn];
vector<int> G[maxn];
struct spaly_tree{
int s[2], fa, val, add, cnt1, cnt_1, fl[2];
ll sum;
}tr[maxn<<1];
void dfs(int x)
{
fir[x] = ++cnt;
tr[cnt].fl[1] = 1;
tr[cnt].val = a[x];
tr[cnt].sum = (ll)a[x];
for(unsigned int i = 0; i < G[x].size(); ++i)
{
int id = G[x][i];
dfs(id);
}
sec[x] = ++cnt;
tr[cnt].fl[0] = 1;
tr[cnt].val = -a[x];
tr[cnt].sum = (ll)(-a[x]);
}
void build(int l, int r, int ff)
{
int mid = (l + r)>>1;
if(l < mid)
build(l, mid - 1, mid);
if(mid < r)
build(mid + 1, r, mid);
if(ff)
tr[ff].s[ff < mid] = mid;
int ls = tr[mid].s[0], rs = tr[mid].s[1];
tr[mid].fa = ff;
tr[mid].cnt1 = tr[ls].cnt1 + tr[rs].cnt1 + tr[mid].fl[1];
tr[mid].cnt_1 = tr[ls].cnt_1 + tr[rs].cnt_1 + tr[mid].fl[0];
tr[mid].sum += tr[ls].sum + tr[rs].sum;
}
void spread(int x)
{
if(tr[x].add)
{
int ls = tr[x].s[0], rs = tr[x].s[1];
if(ls)
{
tr[ls].sum += 1LL * (tr[ls].cnt1 - tr[ls].cnt_1) * tr[x].add;
tr[ls].val += (tr[ls].fl[1]? tr[x].add: -tr[x].add);
tr[ls].add += tr[x].add;
}
if(rs)
{
tr[rs].sum += 1LL * (tr[rs].cnt1 - tr[rs].cnt_1) * tr[x].add;
tr[rs].val += (tr[rs].fl[1]? tr[x].add: -tr[x].add);
tr[rs].add += tr[x].add;
}
tr[x].add = 0;
}
}
void update(int x)
{
int ls = tr[x].s[0], rs = tr[x].s[1];
tr[x].sum = tr[ls].sum + tr[rs].sum + 1LL * tr[x].val;
tr[x].cnt1 = tr[ls].cnt1 + tr[rs].cnt1 + tr[x].fl[1];
tr[x].cnt_1 = tr[ls].cnt_1 + tr[rs].cnt_1 + tr[x].fl[0];
}
void Rotate(int x)
{
int y = tr[x].fa, z = tr[y].fa, k = (tr[y].s[1] == x), w = (tr[z].s[1] == y), son = tr[x].s[k^1];
spread(y);spread(x);
tr[y].s[k] = son;tr[son].fa = y;
tr[x].s[k^1] = y;tr[y].fa = x;
tr[z].s[w] = x;tr[x].fa = z;
update(y);update(x);
}
void Splay(int x, int to)
{
int y, z;
while(tr[x].fa != to)
{
y = tr[x].fa;
z = tr[y].fa;
if(z != to)
Rotate((tr[y].s[0] == x) ^ (tr[z].s[0] == y)? x: y);
Rotate(x);
}
if(!to)
root = x;
}
int Findpre()
{
spread(root);
int now = tr[root].s[0];
while(tr[now].s[1])
{
spread(now);
now = tr[now].s[1];
}
spread(now);
return now;
}
int Findnxt()
{
spread(root);
int now = tr[root].s[1];
while(tr[now].s[0])
{
spread(now);
now = tr[now].s[0];
}
spread(now);
return now;
}
void Remove(int x, int y)
{
Splay(fir[x], 0);
int pre = Findpre();
Splay(sec[x], 0);
int nxt = Findnxt();
Splay(pre, 0);
Splay(nxt, pre);
int now = tr[nxt].s[0];
tr[nxt].s[0] = 0;
update(nxt);update(pre);
Splay(fir[y], 0);
nxt = Findnxt();
tr[now].fa = nxt;
tr[nxt].s[0] = now;
Splay(now, 0);
}
int main()
{
read(n);
for(int i = 1; i < n; ++i)
{
int x;
read(x);
G[x].push_back(i+1);
}
for(int i = 1; i <= n; ++i)
read(a[i]);
cnt = 1;
dfs(1);
cnt ++;
build(1, cnt, 0);
tr[0].s[1] = root = (1 + cnt)>>1;
read(m);
for(int i = 1; i <= m; ++i)
{
char opt[3];
scanf("%s", opt);
if(opt[0] == 'Q')
{
int x;
read(x);
Splay(sec[x], 0);
printf("%lld\n", tr[tr[root].s[0]].sum);
}
else if(opt[0] == 'C')
{
int x, y;
read(x);read(y);
Remove(x, y);
}
else
{
int x, y;
read(x);read(y);
if(!y) continue;
Splay(fir[x], 0);
int pre = Findpre();
Splay(sec[x], 0);
int nxt = Findnxt();
Splay(pre, 0);
Splay(nxt, pre);
int p = tr[nxt].s[0];
tr[p].sum += 1LL * (tr[p].cnt1 - tr[p].cnt_1) * y;
tr[p].val += (tr[p].fl[1]? y: -y);
tr[p].add += y;
update(nxt);update(pre);
}
}
return 0;
}