dp[i][j]表示i節點，已經染色j個節點的最大價值，那就是樹形DP，每個節點做一次揹包

程式碼：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 105;

int n, m, val[N], dp[N][N];
vector<int> g[N];

void dfs(int u, int p) {
    memset(dp[u], 0, sizeof(dp[u]));
    dp[u][1] = val[u];
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == p) continue;
        dfs(v, u);
        for (int j = m; j >= 2; j--) {
            for (int k = 1; k < j; k++) {
                dp[u][j] = max(dp[u][j], dp[u][j - k] + dp[v][k]);
            }
        }
    }
}

int main() {
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; i++) {
            scanf("%d", &val[i]);
            g[i].clear();
        }
        int u, v;
        for (int i = 0; i < n - 1; i++) {
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1, 0);
        printf("%d\n", dp[1][m]);
    }
    return 0;
}