显然每次只会取当前最大的长链。
那么每次直接将所有长链的权值扔到一个堆里面,最后取出(k)次即是最终答案。
写法上类似于树链剖分:
/*
* Author: heyuhhh
* Created Time: 2020/6/11 9:49:25
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << std::endl; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) {
for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
#define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 2e5 + 5;
int n, k;
vector <int> G[N];
int a[N];
ll len[N];
int bson[N];
void dfs(int u, int fa) {
ll Max = 0;
for (auto v : G[u]) if (v != fa) {
dfs(v, u);
if (len[v] > Max) {
Max = len[v];
bson[u] = v;
}
}
len[u] = len[bson[u]] + a[u];
}
int topf[N];
void dfs2(int u, int top, int fa) {
topf[u] = top;
if (bson[u]) {
dfs2(bson[u], top, u);
}
for (auto v : G[u]) {
if (v == fa || v == bson[u]) {
continue;
}
dfs2(v, v, u);
}
}
void run() {
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1, 0);
dfs2(1, 1, 0);
priority_queue <ll> q;
for (int i = 1; i <= n; i++) {
if (topf[i] == i) {
q.push(len[i]);
}
}
ll ans = 0;
while (k--) {
ans += q.top(); q.pop();
}
cout << ans << '
';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}