题解 2020.10.24 考试 T4 模板

题目传送门

题目大意

有一个 (n) 个点组成的树，有 (m) 次操作，每次将 (1 o x) 的路径上每个点都加入一个颜色为 (c) 的小球。但是每个点都有大小限制，即小球个数超过一定量之后就不能再加入了。有 (q) 次查询，问操作完了之后每个点有多少种不同颜色的小球。

思路

stO llsw yyds Orz

以下皆为 llsw yyds 的思路，与本人无关。

我们首先可以整体二分出每一个点在多久会被填满，然后扫描线一波，问题就是如何判断一个子树内不同颜色的个数。这个我们可以直接容斥，就每次加入一个颜色的时候减掉与之前重复的即可。

时间复杂度 (Theta(nlog n))。

( exttt{Code})

#include <bits/stdc++.h>
using namespace std;

#define MAXN 100005
#define Int int

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

vector <int> G[MAXN]; 
int n,m,uni,tmpc[MAXN];

struct Mod{
	int u,c;
}seq[MAXN];

int ind,tur[MAXN],ans[MAXN],lim[MAXN],dfn[MAXN],dep[MAXN],siz[MAXN],par[MAXN][21];
void dfs (int u,int fa){
	dfn[u] = ++ ind,tur[ind] = u,par[u][0] = fa,dep[u] = dep[fa] + 1,siz[u] = 1;
	for (Int i = 1;i <= 20;++ i) par[u][i] = par[par[u][i - 1]][i - 1];
	for (Int v : G[u]) if (v ^ fa) dfs (v,u),siz[u] += siz[v];
}

int getlca (int u,int v){
	if (dep[u] < dep[v]) swap (u,v);
	for (Int dis = dep[u] - dep[v],i = 20;~i;-- i) if (dis >> i & 1) u = par[u][i];
	if (u == v) return u;
	else{
		for (Int i = 20;~i;-- i) if (par[u][i] ^ par[v][i]) u = par[u][i],v = par[v][i];
		return par[u][0];
	}
}

struct Bit_Tree{
	int sum[MAXN];
	int lowbit (int x){return x & (-x);}
	void modify (int x,int v){for (Int i = x;i <= n;i += lowbit (i)) sum[i] += v;}
	int query (int x){int res = 0;for (Int i = x;i;i -= lowbit (i)) res += sum[i];return res;}
	int query (int l,int r){return query (r) - query (l - 1);}
}Tree;

vector <int> qry[MAXN];

void Solve (vector <int> &tmp,int l,int r){
	if (tmp.empty()) return ;
	if (l == r){
		for (Int u : tmp) qry[l].push_back (u); 
		return tmp.clear();
	}
	vector <int> lson,rson;
	int mid = (l + r) >> 1;
	for (Int i = l ? l : 1;i <= mid;++ i) Tree.modify (dfn[seq[i].u],1);
	for (Int u : tmp){
		if (Tree.query (dfn[u],dfn[u] + siz[u] - 1) < lim[u]) rson.push_back (u);  
		else lson.push_back (u); 
	}
	tmp.clear();
	Solve (rson,mid + 1,r);
	for (Int i = l ? l : 1;i <= mid;++ i) Tree.modify (dfn[seq[i].u],-1);
	Solve (lson,l,mid); 
}

set <int> colS[MAXN];

void Insert (set <int> &S,int u){
	auto res = S.insert (dfn[u]);
	if (!res.second) return ;
	Tree.modify (dfn[u],1); 
	auto it = res.first;int p = -1,q = -1;
	if (it != S.begin()) Tree.modify (dfn[getlca (u,p = tur[*-- it])],-1),++ it;
	if (++ it != S.end()) Tree.modify (dfn[getlca (u,q = tur[*it])],-1);
	if (~q && ~p) Tree.modify (dfn[getlca (p,q)],1); 
}

signed main(){
	read (n);
	for (Int i = 2,u,v;i <= n;++ i) read (u,v),G[u].push_back (v),G[v].push_back (u);
	for (Int i = 1;i <= n;++ i) read (lim[i]);
	read (m);
	for (Int i = 1;i <= m;++ i) read (seq[i].u,seq[i].c),tmpc[i] = seq[i].c;
	sort (tmpc + 1,tmpc + m + 1),uni = unique (tmpc + 1,tmpc + m + 1) - tmpc - 1;
	for (Int i = 1;i <= m;++ i) seq[i].c = lower_bound (tmpc + 1,tmpc + m + 1,seq[i].c) - tmpc;
	dfs (1,0);
	vector <int> tmp;for (Int i = 1;i <= n;++ i) tmp.push_back (i);
	Solve (tmp,0,m);
	for (Int i = 1;i <= m;++ i){
		Insert (colS[seq[i].c],seq[i].u);
		for (Int u : qry[i]) ans[u] = Tree.query (dfn[u],dfn[u] + siz[u] - 1); 
	} 
	int q;read (q);
	for (Int i = 1,x;i <= q;++ i) read (x),write (ans[x]),putchar ('
');
	return 0;
}