题目链接：hihocoder1055

【分析】用背包的思想进行树形DP，注意如果要选择该子树根是必须要选择的，否则无法和下面的连通。

【AC CODE】0ms

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <map>
//#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 100+10, MAXM = 100*2+10;
int head[MAXN], tol, nxt[MAXM], to[MAXM];
int val[MAXN], dp[MAXN][MAXN], m, num[MAXN];

inline void add_edge(int u, int v)
{
	nxt[tol] = head[u], to[tol] = v;
	head[u] = tol++;
}
void dfs(int u, int fa)
{
	repe(i,1,m) dp[u][i] = val[u];
	for(int i = head[u]; ~i; i = nxt[i])
	{
		int v = to[i];
		if(fa == v) continue;
		dfs(v,u);
		num[u] += num[v];
		int mx = min(num[v],m);
		int tmp[MAXN];
		memcpy(tmp,dp[u],sizeof(dp[u]));
		repe(j,1,mx)//v子树取j个点
		{
			repe(k,j+1,m)//u树取k个
				tmp[k] = max(tmp[k], dp[u][k-j]+dp[v][j]);
		}
		repe(i,0,m) dp[u][i] = max(dp[u][i], tmp[i]);
	}
	num[u]++;
}
int main()
{
#ifdef SHY
	freopen("d:\\1.txt", "r", stdin);
#endif
	int n;
	while(~scanf("%d %d%*c", &n, &m))
	{
		repe(i,1,n) scanf("%d%*c", &val[i]);
		clc(head,-1);
		tol = 0;
		rep(i,1,n)
		{
			int u,v;
			scanf("%d %d%*c", &u, &v);
			add_edge(u,v);
			add_edge(v,u);
		}
		clc(dp,0);
		clc(num,0);
		dfs(1,-1);
		printf("%d\n", dp[1][m]);
	}
	return 0;
}