[CF960H]Santa's Gift

2020-03-20
#树 #线段树 #树链剖分
Codeforces

题意

给出一棵有$n$个节点的树，并染上$m$种颜色，给出$\{b_i\}$和$C$

修改操作：将一个节点的颜色改变

询问操作：求$\frac{\sum_{i=1}^n (S_{i}\cdot b_x-C)^2}{n}$，其中S是根节点为i的子树中与颜色为x的数量

$n,m,Q\leq 50000​$

题解

先把式子拆开看看

可以注意到$\sum_{i=1}^n S_i$等价于所有颜色为x的节点的深度和，可以直接维护

剩下区间平方和，树链剖分+线段树强行维护即可

调试记录

区间tg下传的时候傻掉了

第一次觉得每次+1的贡献都一样，实际上S在变啊

还有没写tg的继承

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#include <cstdio>
#include <algorithm>
#define LL long long
#define int long long
const int maxn = 5e4 + 5;
using namespace std;
struct E{
	int to, nxt;
}e[maxn];
int head[maxn], tot = 0;
void addedge(int u, int v){
	e[++tot].to = v, e[tot].nxt = head[u];
	head[u] = tot;
}
int f[maxn], son[maxn], sz[maxn], dep[maxn];
void dfs1(int cur, int fa){
	f[cur] = fa; sz[cur] = 1; dep[cur] = dep[fa] + 1;
	int Max = -1;
	for (int i = head[cur]; i; i = e[i].nxt){
		int v = e[i].to;
		dfs1(v, cur);
		sz[cur] += sz[v];
		if (Max < sz[v]) Max = sz[v], son[cur] = v;
	}
}
int top[maxn], id[maxn], tdx = 0;
void dfs2(int cur, int topf){
	top[cur] = topf;
	id[cur] = ++tdx;
	if (!son[cur]) return;
	dfs2(son[cur], topf);
	for (int i = head[cur]; i; i = e[i].nxt)
		if (e[i].to != son[cur]) dfs2(e[i].to, e[i].to);
} int n, m, Q, c[maxn], b[maxn], C;
LL _abs(LL x){return x < 0 ? -x : x;}
struct T{
	struct A{
		int ls, rs;
		LL v, v2, tg;
	}a[maxn * 200]; int ttot;
	#define LS a[a[cur].ls]
	#define RS a[a[cur].rs]
	void pushdown(int cur, int l, int r){
		if (a[cur].tg == 0) return;
		int mid = l + r >> 1;
		if (a[cur].ls == 0) a[cur].ls = ++ttot;
		LS.v2 += 2ll * a[cur].tg * LS.v + 1ll * a[cur].tg * a[cur].tg * (mid - l + 1); 
		LS.v += 1ll * a[cur].tg * (mid - l + 1);
		if (a[cur].rs == 0) a[cur].rs = ++ttot;
		RS.v2 += 2ll * a[cur].tg * RS.v + 1ll * a[cur].tg * a[cur].tg * (r - mid);
		RS.v += 1ll * a[cur].tg * (r - mid);
		LS.tg += a[cur].tg;
		RS.tg += a[cur].tg;
		a[cur].tg = 0;
	}
	void upd(int &cur, int l, int r, int L, int R, int k){
		if (l > R || r < L) return;
		if (!cur) cur = ++ttot;
		if (l >= L && r <= R){
			a[cur].v2 += 2ll * k * a[cur].v + 1ll * (r - l + 1) * k * k;
			a[cur].v += 1ll * k * (r - l + 1);
			a[cur].tg += k;
			return;
		}
		pushdown(cur, l, r);
		int mid = l + r >> 1;
		upd(a[cur].ls, l, mid, L, R, k);
		upd(a[cur].rs, mid + 1, r, L, R, k);
		a[cur].v = LS.v + RS.v;
		a[cur].v2 = LS.v2 + RS.v2;
	}
}t;
void urange(int col, int u, int v, int k){
	while (top[u] != top[v]){
		if (dep[top[u]] > dep[top[v]]) swap(u, v);
		t.upd(col, 1, n, id[top[v]], id[v], k);
		v = f[top[v]];
	}
	if (dep[u] > dep[v]) swap(u, v);
	t.upd(col, 1, n, id[u], id[v], k);
}
LL sdep[maxn];
signed main(){
	scanf("%lld%lld%lld%lld", &n, &m, &Q, &C); t.ttot = m;
	for (int i = 1; i <= n; i++) scanf("%lld", c + i);
	for (int fa, i = 2; i <= n; i++)
		scanf("%lld", &fa), addedge(fa, i);
	for (int i = 1; i <= m; i++) scanf("%lld", b + i);
	dfs1(1, 0); dfs2(1, 1);
	for (int i = 1; i <= n; i++)
		urange(c[i], 1, i, 1), sdep[c[i]] += dep[i];
	while (Q--){
		int opt, x, y; scanf("%lld", &opt);
		if (opt == 1){
			scanf("%lld%lld", &x, &y);
//			if (y == c[x]) continue;
			urange(c[x], 1, x, -1); sdep[c[x]] -= dep[x];
			urange(c[x] = y, 1, x, 1); sdep[c[x]] += dep[x];
		}
		if (opt == 2){
			scanf("%lld", &x);
			double ans = C * C;
			ans += 1.0 * b[x] * b[x] / n * t.a[x].v2;
			ans -= 2.0 * C * b[x] / n * sdep[x];
			printf("%.8lf\n", ans);
		}
//		for (int i = 1; i <= m; i++)
//			printf("%d ", t.Query(i, 1, n, 1, n));
//		puts("\n---------");
	}
	return 0;
}