题解：P10641 BZOJ3252 攻略

我让 cz 搬这道题，cz 给搬了，于是来写个题解（

考虑一个朴素的贪心：每次选择一个到根路径价值和最大的叶子，将价值和累加进答案，并把这条链价值清零。

这个贪心的正确性显然（可以交换法证明），很容易用数据结构维护做到 \(O(n\log^2 n)\) 或 \(O(n\log n)\)。

但是这样太不优美了，而且数据结构比较难写，于是考虑一个更优的做法。

定义路径权值为路径的价值和，对树进行长链剖分。容易发现，贪心的算法流程等价于每次选择权值最大的一条长链，因此答案即为前 \(k\) 大的长链的权值之和，可以做到 \(O(n)\)。

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
    uniform_int_distribution<int> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

template<int mod>
inline unsigned int down(unsigned int x) {
    return x >= mod ? x - mod : x;
}

template<int mod>
struct Modint {
    unsigned int x;
    Modint() = default;
    Modint(unsigned int x) : x(x) {}
    friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
    friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
    friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
    friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
    friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
    friend Modint operator/(Modint a, Modint b) {return a * ~b;}
    friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
    friend Modint operator~(Modint a) {return a ^ (mod - 2);}
    friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
    friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
    friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
    friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
    friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
    friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
    friend Modint& operator++(Modint& a) {return a += 1;}
    friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
    friend Modint& operator--(Modint& a) {return a -= 1;}
    friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
    friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
    friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};

const int N = 2e5 + 5;

int n, k, a[N], fa[N], son[N], vis[N];
ll val[N], ans;
vector<int> e[N];
vector<ll> v;

void dfs(int u, int f) {
    fa[u] = f;
    for(int v : e[u]) {
        if(v == f) continue;
        dfs(v, u);
        if(val[v] > val[son[u]]) son[u] = v;
    }
    val[u] = val[son[u]] + a[u];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n >> k;
    rep(i, 1, n) cin >> a[i];
    rep(i, 1, n - 1) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1, 0);
    rep(i, 1, n) vis[son[i]] = 1;
    rep(i, 1, n) if(!vis[i]) v.push_back(val[i]);
    nth_element(v.begin(), v.begin() + k, v.end(), greater<ll>());
    rep(i, 0, min(k, (int)v.size()) - 1) ans += v[i];
    cout << ans << endl;
    return 0;
}