胡闹 方差

题目大意

维护一颗$n\le10^5$个点的树,每个点有颜色$c_i\le10^5$。

初始时只有$1$号节点。

三种操作:

  1. 加入一个点,给定它的编号、颜色和父亲。
  2. 询问路径$(x,y)$上的所有点。
  3. 询问点$x$的子树内的所有点(以$1$为根)。

询问的意思是说,把所有的点的颜色去重求方差。

解析

我们可以离线,那么完全可以把两个操作分开来处理。

操作$2$直接树上莫队。

操作$3$直接带修莫队。

完了。

Code

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#define DREP(i, s, e) for(int i = s; i >= e ;i--)
#define REP(i, s, e) for(int i = s; i <= e ;i++)

#define DEBUG fprintf(stderr, "Passing [%s] in Line %d\n", __FUNCTION__, __LINE__)
#define chkmax(a, b) a = max(a, b)
#define chkmin(a, b) a = min(a, b)

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 1e5 + 10, maxm = maxn;

int bg[maxn], ne[maxm << 1], to[maxm << 1], e;
inline void add(int x, int y)
{
e++;
to[e] = y;
ne[e] = bg[x];
bg[x] = e;
}

template <typename T> inline T read()
{
T ans(0), p(1);
char c = getchar();
while (!isdigit(c))
{
if (c == '-') p = -1;
c = getchar();
}
while (isdigit(c))
{
ans = ans * 10 + c - 48;
c = getchar();
}
return ans * p;
}

int n = 1, m;
int val[maxm];

struct opt
{
int type, a, b;
}o[maxm];

int st[maxn], ed[maxn], back[maxn<<1], dfn[maxn], dfs_clock, ww[maxn], fa[maxn], top[maxn], siz[maxn], hvy[maxn], depth[maxn];
void dfs1(int x)
{
back[st[x] = ++dfs_clock] = x;
siz[x] = 1;
for (int i = bg[x]; i ; i = ne[i])
{
depth[to[i]] = depth[x] + 1;
dfs1(to[i]);
siz[x] += siz[to[i]];
if (siz[to[i]] > siz[hvy[x]]) hvy[x] = to[i];
}
back[ed[x] = ++dfs_clock] = x;
}
void dfs2(int x, int y)
{
ww[dfn[x] = ++dfs_clock] = val[x];
top[x] = y;
if (hvy[x]) dfs2(hvy[x], y);
for (int i = bg[x]; i ; i = ne[i])
if (to[i] ^ hvy[x]) dfs2(to[i], to[i]);
}
inline int lca(int x, int y)
{
while (top[x] ^ top[y])
{
if (depth[top[x]] < depth[top[y]]) swap(x, y);
x = fa[top[x]];
}
return depth[x] < depth[y] ? x : y;
}
int belong[maxn<<1];
double ans[maxn];

int cnt[maxn], l, r, T;
long long sigma, sigma2, N;
#define bar ((double)sigma / N)
#define S2 ((sigma2 + bar * bar * N - bar * sigma * 2) / N)

inline void add(long long c) {if (c && !cnt[c]++) sigma += c, sigma2 += c * c, N++;}
inline void del(long long c) {if (c && !--cnt[c]) sigma -= c, sigma2 -= c * c, N--;}
struct query
{
int l, r, t, id, spj;
query(int _l = 0, int _r = 0, int _id = 0, int _spj = 0, int _t = 0) : l(_l), r(_r), id(_id), spj(_spj), t(_t){}
}q[maxm];
inline bool operator < (query A, query B) {return belong[A.l] < belong[B.l] || (belong[A.l] == belong[B.l] && belong[A.r] < belong[B.r]) || (belong[A.l] == belong[B.l] && belong[A.r] == belong[B.r] && A.t < B.t);}
namespace path
{
bool vis[maxn];
inline void opt(int x)
{
if (vis[x]) del(val[x]);
else add(val[x]);
vis[x] ^= 1;
}
int Q;
inline void work()
{
REP(i, 1, m)
if (o[i].type == 2)
{
int x(o[i].a), y(o[i].b), l(lca(x, y));
if (st[x] > st[y]) swap(x, y);
if (l ^ x && l ^ y) q[++Q] = query(ed[x], st[y], i, l);
else q[++Q] = query(st[x], st[y], i);
}
const int block_siz = ceil(sqrt(2 * n));
REP(i, 1, n+n) belong[i] = i / block_siz;
sort(q + 1, q + 1 + Q);
REP(i, 1, Q)
{
while (l < q[i].l) opt(back[l++]);
while (l > q[i].l) opt(back[--l]);
while (r < q[i].r) opt(back[++r]);
while (r > q[i].r) opt(back[r--]);
if (q[i].spj) opt(q[i].spj);
ans[q[i].id] = S2;
if (q[i].spj) opt(q[i].spj);
}
}
}
int ade02[maxn], cur;
bool inside[maxn];
inline void timego()
{
int x = ade02[++T];
inside[dfn[x]] = 1;
if (l <= dfn[x] && dfn[x] <= r) add(val[x]);
}
inline void timeback()
{
int x = ade02[T--];
inside[dfn[x]] = 0;
if (l <= dfn[x] && dfn[x] <= r) del(val[x]);
}
inline void Add(int x) {if (inside[x]) add(ww[x]);}
inline void Del(int x) {if (inside[x]) del(ww[x]);}
inline void output()
{
REP(i,1,100000)if(cnt[i])printf("%d ",i);putchar(10);
printf("%d %lld %lld %lf\n", N, sigma, sigma2, S2);
}
namespace subtree
{
int Q;
void work()
{
inside[1] = 1;
REP(i, 1, m)
if (o[i].type == 1) ade02[++cur] = o[i].a;
else if (o[i].type == 3) q[++Q] = query(dfn[o[i].a], dfn[o[i].a] + siz[o[i].a] - 1, i, 0, cur);
if (!Q) return;
const int block_siz = ceil(pow(1. * n * Q, 1. / 3));
REP(i, 1, n) belong[i] = i / block_siz;
sort(q + 1, q + 1 + Q);
sigma = sigma2 = N = l = r = 0;
memset(cnt, 0, sizeof(cnt));
REP(i, 1, Q)
{
while (T < q[i].t) timego();
while (T > q[i].t) timeback();
while (l < q[i].l) Del(l++);
while (l > q[i].l) Add(--l);
while (r < q[i].r) Add(++r);
while (r > q[i].r) Del(r--);
ans[q[i].id] = S2;
}
}
}

signed main()
{
#ifdef CraZYali
freopen("A.in", "r", stdin);
freopen("A.out", "w", stdout);
#endif
cin >> m >> val[1];
REP(i, 1, m)
{
int type(read<int>()), a(read<int>());
if (type == 1)
{
++n;
o[i] = opt{type, a};
val[a] = read<int>();
add(fa[a] = read<int>(), a);
}
else if (type == 2) o[i] = opt{type, a, read<int>()};
else o[i] = opt{type, a};
}
dfs1(1);dfs_clock = 0;
dfs2(1, 1);
path::work();
subtree::work();
REP(i, 1, m)
if (o[i].type > 1) printf("%.10lf\n", ans[i]);
return 0;
}

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