CodeForces 293E Close Vertices

Description
You've got a weighted tree, consisting of n vertices. Each edge has a non-negative weight. The length of the path between any two vertices of the tree is the number of edges in the path. The weight of the path is the total weight of all edges it contains.
Two vertices are close if there exists a path of length at most l between them and a path of weight at most w between them. Count the number of pairs of vertices v, u(v < u), such that vertices v and u are close.
Input
The first line contains three integers n, l and w(1 ≤ n ≤ 105, 1 ≤ l ≤ n, 0 ≤ w ≤ 109). The next n - 1 lines contain the descriptions of the tree edges. The i-th line contains two integers pi, wi(1 ≤ pi < (i + 1), 0 ≤ wi ≤ 104), that mean that the i-th edge connects vertex (i + 1) and pi and has weight wi.
Consider the tree vertices indexed from 1 to n in some way.
Output
Print a single integer — the number of close pairs.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Sample Input
Input

4 4 6
1 3
1 4
1 3

Output

4

Input

6 2 17
1 3
2 5
2 13
1 6
5 9

Output
9
나무를 나누어 치료한 다음에 그 안에는 2차원 대조 처리가 있다. 이것은 내가 쓴 편차 문제에 대한 처리 방법을 어떻게 참고할 수 있습니까?
여기는 트리 그룹 유지보수를 직접 정렬할 수 있습니다. 제가 직접 쓴 cdq분할을 편리하게 하기 위해서입니다.

#include<queue>
#include<cstdio>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int INF = 0x7FFFFFFF;
const int maxn = 2e5 + 10;
int n, l, w, x, y;

struct Tree
{
	int ft[maxn], nt[maxn], u[maxn], v[maxn], sz;
	int mx[maxn], ct[maxn], vis[maxn], t;
	struct point
	{
		int x, y, z;
		point(int x = 0, int y = 0, int z = 0) :x(x), y(y), z(z) {};
		bool operator<(const point&a)const
		{
			return x == a.x ? y == a.y ? z < a.z : y < a.y : x < a.x;
		}
	}a[maxn], b[maxn];

	void clear(int n)
	{
		mx[sz = 0] = INF;
		for (int i = 1; i <= n; i++) ft[i] = -1, vis[i] = 0;
	}
	void AddEdge(int x, int y, int z)
	{
		u[sz] = y;	v[sz] = z;	nt[sz] = ft[x];	ft[x] = sz++;
		u[sz] = x;	v[sz] = z;	nt[sz] = ft[y]; ft[y] = sz++;
	}
	int dfs(int x, int fa, int sum)
	{
		int y = mx[x] = (ct[x] = 1) - 1;
		for (int i = ft[x]; i != -1; i = nt[i])
		{
			if (vis[u[i]] || u[i] == fa) continue;
			int z = dfs(u[i], x, sum);
			ct[x] += ct[u[i]];
			mx[x] = max(mx[x], ct[u[i]]);
			y = mx[y] < mx[z] ? y : z;
		}
		mx[x] = max(mx[x], sum - ct[x]);
		return mx[x] < mx[y] ? x : y;
	}
	void get(int x, int fa, int dep, int len)
	{
		a[t++] = point(dep, len, 0);
		a[t++] = point(l - dep, w - len, 1);
		for (int i = ft[x]; i != -1; i = nt[i])
		{
			if (u[i] == fa || vis[u[i]]) continue;
			get(u[i], x, dep + 1, len + v[i]);
		}
	}
	LL merge(int l, int r)
	{
		if (l == r) return 0;
		int m = l + r >> 1;
		LL ans = merge(l, m) + merge(m + 1, r);
		for (int i = l, j = l, k = m + 1, ct = 0; i <= r; i++)
		{
			if (j <= m && (a[j].y < a[k].y || a[j].y == a[k].y&&a[j].z <= a[k].z || k > r))
			{
				b[i] = a[j++];	ct += b[i].z ^ 1;
			}
			else
			{
				b[i] = a[k++];	ans += b[i].z*ct;
			}
		}
		for (int i = l; i <= r; i++) a[i] = b[i];
		return ans;
	}
	LL find(int x, int dep, int len)
	{
		LL ans = t = 0;
		get(x, -1, dep, len);
		sort(a, a + t);
		for (int i = 0; i < t; i++)
		{
			if (a[i].x + a[i].x <= l&&a[i].y + a[i].y <= w) ans += a[i].z ^ 1;
		}
		return merge(0, t - 1) - ans >> 1;
	}
	LL work(int x, int sum)
	{
		int y = dfs(x, -1, sum);
		LL ans = find(y, 0, 0);	vis[y] = 1;
		for (int i = ft[y]; i != -1; i = nt[i])
		{
			if (vis[u[i]]) continue;
			ans -= find(u[i], 1, v[i]);
			ans += work(u[i], ct[u[i]] > ct[y] ? sum - ct[y] : ct[u[i]]);
		}
		return ans;
	}
}solve;

int main()
{
	while (scanf("%d%d%d", &n, &l, &w) != EOF)
	{
		solve.clear(n);
		for (int i = 1; i < n; i++)
		{
			scanf("%d%d", &x, &y);
			solve.AddEdge(i + 1, x, y);
		}
		printf("%I64d
", solve.work(1, n));
	}
	return 0;
}