A - Coloring Roads Gym - 102059A

Posted 2020-12-16 lordxx

用lct中的access操作来维护链。因为每一次更新到根节点所以不用split操作。
注意下传标记和统计的时候影响。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<‘0‘ || ch>‘9‘) { if (ch == ‘-‘)f = -1; ch = getchar(); }
	while (ch >= ‘0‘ && ch <= ‘9‘) { x = x * 10 + ch - ‘0‘; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 4e5 + 10;
int n, C, Q;
int cnt[N], num[N];
struct SPLAY {
	int ch[N][2];
	int val[N];
	int sz[N];
	int fa[N]; int lazy[N];
	void change(int x, int c)
	{
		val[x] = lazy[x] = c;
	}
	void pushdown(int x)
	{
		if (lazy[x])
		{
			change(ch[x][0], lazy[x]);
			change(ch[x][1], lazy[x]);
			lazy[x] = 0;
		}
	}
	void pushall(int x)
	{
		if (!isroot(x))pushall(fa[x]);
		pushdown(x);
	}
	int ident(int u, int f)
	{
		return ch[f][1] == u;
	}
	void pushup(int root)
	{
		sz[root] = root > n ? 1 : 0;
		sz[root] += sz[ch[root][0]];
		sz[root] += sz[ch[root][1]];
	}
	int isroot(int x)
	{
		return (ch[fa[x]][0] != x && ch[fa[x]][1] != x);
	}
	void connect(int x, int f, int is)
	{
		ch[f][is] = x;
		fa[x] = f;
	}
	void rotate(int x)
	{
		int f = fa[x]; int ff = fa[f];
		int k = ident(x, f);
		connect(ch[x][k ^ 1], f, k);
		fa[x] = ff;
		if (!isroot(f))ch[ff][ident(f, ff)] = x;
		connect(f, x, k ^ 1);
		pushup(f); pushup(x);
	}
	void splay(int x)
	{
		pushall(x);
		while (!isroot(x))
		{
			int f = fa[x]; int ff = fa[f];
			if (!isroot(f))
				ident(f, ff) ^ ident(x, f) ? rotate(x) : rotate(f);
			rotate(x);
		}
	}
	void cal(int x, int color, int s)
	{
		num[cnt[x]]--;
		cnt[x] -= s;
		num[cnt[x]]++;
		num[cnt[color]]--;
		cnt[color] += s;
		num[cnt[color]]++;
	}
	void access(int x, int color)
	{
		for (int y = 0; x; x = fa[x])
		{
			splay(x);
			cal(val[x], color, sz[x] - sz[ch[x][1]]);
			ch[x][1] = y;
			y = x;
			change(x, color);
			pushup(x);
		}
	}
}T;
vector<int>vec[N];
int dfn[N], cnt_d;
void dfs(int u, int fa)
{
	dfn[u] = ++cnt_d;
	for (auto k : vec[u])
	{
		if (k == fa)continue;
		dfs(k, u);
		T.fa[n + dfn[k]] = u;
		T.fa[k] = n + dfn[k];
	}
}
int main()
{
	n = read(); C = read(); Q = read();
	int u, v;
	upd(i, 1, n - 1)
	{
		u = read(), v = read();
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	dfs(1, 0);
	upd(i, 1, n + cnt_d)
		T.pushup(i);
	cnt[0] = n - 1;
	num[n - 1] = 1;
	num[0] += C;
	int c, m;
	while (Q--)
	{
		u = read(); c = read(), m = read();
		T.access(u, c);
		printf("%d
", num[m] - (cnt[0] == m));
	}
	return 0;
}