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

树分治，然后里面是二维点对的处理，这个怎么做可以参考之前我写的关于偏序的问题的处理方法。

这里可以直接排序加树状数组维护，为了方便一点我直接写的cdq分治。

#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\n", solve.work(1, n));
}
return 0;
}