/*
This problem can be found at https://content-available-to-author-only.uz/problem/view/IOI11_race.
SOLUTION
Consider the problem of finding a path of length K containing minimum edges THAT GOES A FIXED NODE x.
This can be done in O(Size of Tree) time. Once this is done, we can remove the node x and edges connected to it. Then we can solve the problem for the remaining connected components.
The best choice of x is the centroid of the tree.
*/
#include "race.h"
#include "bits/stdc++.h"
using namespace std;
const int mxN = 2e5L + 10;
const int mxK = 1e6L + 10;
int n, k;
vector<pair<int, int>> adj[mxN];
int ans = mxN;
int sz[mxN];
bool proc[mxN] = {};
int dp[mxK];
int get_sz(int x, int p) {
sz[x] = 1;
for(auto [y, w] : adj[x]) {
if(proc[y] || y == p) continue;
sz[x] += get_sz(y, x);
}
return sz[x];
}
int get_cen(int x, int p, int tot) {
for(auto [y, w] : adj[x]) {
if(proc[y] || y == p || sz[y] * 2 <= tot) continue;
return get_cen(y, x, tot);
}
return x;
}
void dfs(int x, int p, int dep, int dis, int flag) {
if(dis > k) return;
if(flag == 0) {
if(dp[k - dis] != mxN) ans = min(ans, dep + dp[k - dis]);
}
else if(flag == 1) {
dp[dis] = min(dp[dis], dep);
}
else if(flag == 2) {
dp[dis] = mxN;
}
for(auto [y, w] : adj[x]) {
if(proc[y] || y == p) continue;
dfs(y, x, dep + 1, dis + w, flag);
}
}
void solve(int x = 1, int p = -1) {
int c = get_cen(x, p, get_sz(x, p));
dp[0] = 0;
for(auto [d, w] : adj[c]) {
if(proc[d]) continue;
dfs(d, c, 1, w, 0);
dfs(d, c, 1, w, 1);
}
for(auto [d, w] : adj[c]) {
if(proc[d]) continue;
dfs(d, c, 1, w, 2);
}
dp[0] = mxN;
proc[c] = 1;
for(auto [d, w] : adj[c]) {
if(proc[d]) continue;
solve(d, c);
}
}
int best_path(int N, int K, int H[][2], int L[])
{
for(int i = 0; i < mxK; ++i) dp[i] = mxN;
n = N; k = K;
for(int i = 0; i < n - 1; ++i) {
adj[H[i][0] + 1].emplace_back(H[i][1] + 1, L[i]);
adj[H[i][1] + 1].emplace_back(H[i][0] + 1, L[i]);
}
solve();
return (ans > n - 1 ? -1 : ans);
}
