【gym 102861E】:Party Company 树状数组 + 离线

传送门

分析

不想翻译了。。
这道题跟牛客多校那道题很类似，但是因为做法不同就导致了这两个题的做法完全不同了
我们去离线处理每一个询问，找到这个询问能往上走的最高点，然后在树状数组对应的位置打上标记，然后向下$dfs$即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n,a[N],m;
int tr[N];
int fa[N][25], mx[N][25];
int h[N],e[M],ne[M],idx;
VI query[N];
int ans[N];

int lowbit(int x){
    return x & -x;
}

void ad(int x,int y){
	ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void add(int x,int c){
    for(int i = x;i < N;i += lowbit(i)) tr[i] += c;
}

int sum(int x){
    int res = 0;
    for(int i = x;i;i -= lowbit(i)) res += tr[i];
    return res;
}

int ans_up(int x, int k) {
	for (int i = 20; i >= 0; i--) {
		if (fa[x][i] && mx[x][i] <= k) x = fa[x][i];
	}
	return x;
}


void dfs(int u, int f) {
	fa[u][0] = f;
	if (u != 1) mx[u][0] = a[f];
	else mx[u][0] = INF;
	for (int k = 1; k <= 20; k++) {
		fa[u][k] = fa[fa[u][k - 1]][k - 1];
		mx[u][k] = max(mx[u][k - 1], mx[fa[u][k - 1]][k - 1]);
	}
	for (int i = h[u]; ~i; i = ne[i]) {
		int j = e[i];
		if (j == f) continue;
		dfs(j, u);
	}
}

void dfs_(int u,int f){
	for(int p:query[u]) add(p,1);
	ans[u] 	= sum(a[u]);
	for(int i = h[u];~i;i = ne[i]){
		int j = e[i];
		if(j == f) continue;
		dfs_(j,u);
	}
	for(int p:query[u]) add(p,-1);
}

int main() {
	memset(h,-1,sizeof h);
	read(n),read(m);
	for(int i = 1;i <= n;i++){
		int x,y;
		read(x),read(y);
		if(i == 1){
			ad(1,0);
			a[1] = x;
		}	
		else{
			ad(y,i);
			a[i] = x;
		}
	}
	dfs(1,0);
	for(int i = 1;i <= m;i++){
		int x,y,z;
		read(x),read(y),read(z);
		x = ans_up(x,z);
		query[x].pb(y);
	}
	dfs_(1,0);
	for(int i = 1;i <= n;i++) printf("%d ",ans[i]);
	return 0;
}