分析：一個貪心的想法是每次找到根的點權和最大的點進行操作，關鍵是怎麽維護.每次找最大值，修改後會對這條鏈上每個點的子樹上的點造成影響，可以用線段樹來維護.找最大值就是區間求最大值嘛，對子樹進行操作利用dfs序維護一下就好了.記錄一下最大值的位置，每次從這個位置向上跳並對它的子樹進行修改直到當前點的點權為0.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 200010
 
;
typedef long long ll;

int n, k, a[maxn], fa[maxn],head[maxn], to[maxn], nextt[maxn],tot = 1, d[maxn], l[maxn], r[maxn], pos[maxn], dfs_clock;
int maxx[maxn << 2], tag[maxn << 2], p[maxn << 2];
ll ans;

void add(int x, int y)
{
    to[tot] = y;
    nextt[tot] = head[x];
    head[x] 
 
= tot++;
}

void dfs(int u, int from, int dist)
{
    d[u] = dist;
    l[u] = ++dfs_clock;
    pos[dfs_clock] = u;
    for (int i = head[u]; i; i = nextt[i])
    {
        int v = to[i];
        if (v != from)
        {
            dfs(v, u, dist + a[v]);
            fa[v] = u;
        }
    }
    r[u] 
 
= dfs_clock;
}

void pushup(int o)
{
    if (maxx[o * 2] > maxx[o * 2 + 1])
    {
        maxx[o] = maxx[o * 2];
        p[o] = p[o * 2];
    }
    else
    {
        maxx[o] = maxx[o * 2 + 1];
        p[o] = p[o * 2 + 1];
    }
}

void build(int o, int l, int r)
{
    if (l == r)
    {
        maxx[o] = d[pos[l]];
        p[o] = pos[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(o * 2, l, mid);
    build(o * 2 + 1, mid + 1, r);
    pushup(o);
}

void pushdown(int o)
{
    if (tag[o] != 0)
    {
        tag[o * 2] += tag[o];
        tag[o * 2 + 1] += tag[o];
        maxx[o * 2] += tag[o];
        maxx[o * 2 + 1] += tag[o];
        tag[o] = 0;
    }
}

void update(int o, int l, int r, int x, int y,int v)
{
    if (x <= l && r <= y)
    {
        maxx[o] += v;
        tag[o] += v;
        return;
    }
    pushdown(o); 
    int mid = (l + r) >> 1;
    if (x <= mid)
        update(o * 2, l, mid, x, y, v);
    if (y > mid)
        update(o * 2 + 1, mid + 1, r, x, y, v);
    pushup(o);
}

void change(int x)
{
    while (a[x])
    {
        update(1, 1, n, l[x], r[x], -a[x]);
        a[x] = 0;
        x = fa[x];
    }
}

int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for (int i = 1; i < n; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }
    dfs(1, 0, a[1]);
    build(1, 1, n);
    for (int i = 1; i <= k; i++)
    {
        ans += maxx[1];
        change(p[1]);
    }
    printf("%lld\n", ans);

    return 0;
}

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