P6326

Prob

树上选择一些物品，有多个，要求选择的所有物品成一个连通块，求出最大的价值。

Sol

点分治，考虑强制选择某个点 $u$ 的答案，那么就以 $u$ 为根做多重背包，并且特判必须选 $u$ ，对于 $n$ 个点都做一遍强制选，那么就可以取到最优解。可以证明点分治也同样可以取到了所有连通块，简单证明一下：

设点分治所有选择的重心按顺序分别为 $p_{i}$ ，设当前连通块为 $s$ ，那么：

对于每一个连通块 $s$ 会被满足以下条件的 $p_{i}$ 统计到：

$p_{j} (j < i) \notin s$ 。
$p_{i} \in s$ 。

显然每个连通块只有且只有一个 $i$ 满足条件，也就是说每个连通块被不重不漏的统计了一遍。

证明有误请大神们在评论区留言/kel

Code

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define pf push_front
#define desktop "C:\\Users\\incra\\Desktop\\"
#define IOS ios :: sync_with_stdio (false),cin.tie (0),cout.tie (0)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair <int,int> PII;
const int dx[] = {1,0,-1,0},dy[] = {0,-1,0,1};
template <typename T1,typename T2> bool tomax (T1 &x,T2 y) {
	if (y > x) return x = y,true;
	return false;
}
template <typename T1,typename T2> bool tomin (T1 &x,T2 y) {
	if (y < x) return x = y,true;
	return false;
}
LL power (LL a,LL b,LL p) {
	LL ans = 1;
	while (b) {
		if (b & 1) ans = ans * a % p;
		a = a * a % p;
		b >>= 1;
	}
	return ans;
}
int fastio = (IOS,0);
#define endl '\n'
#define puts(s) cout << (s) << endl
const int N = 510,M = 4010;
int n,m;
vector <int> g[N];
int rt;
bool del[N];
int v[N],w[N],d[N];
int ww[N],vv[N];
int ans;
int dfn[N],num[N],last[N],timestamp;
int f[N][M];
int get_size (int u,int fa) {
    int s = 1;
    for (int v : g[u]) {
        if (v == fa || del[v]) continue;
        s += get_size (v,u);
    }
    return s;
}
int get_root (int u,int fa,int tot) {
    int maxx = 0,s = 1;
    for (int v : g[u]) {
        if (v == fa || del[v]) continue;
        int t = get_root (v,u,tot);
        tomax (maxx,t);
        s += t;
    }
    tomax (maxx,tot - s);
    if (maxx <= tot / 2) rt = u;
    return s;
}
void DFS (int u,int fa) {
    num[dfn[u] = ++timestamp] = u;
    for (int v : g[u]) {
        if (v == fa || del[v]) continue;
        DFS (v,u);
    }
    last[u] = timestamp;
}
void calc (int u) {
    timestamp = 0;
    DFS (u,-1);
    for (int i = timestamp;i >= 1;i--) {
        int s = d[num[i]] - 1;
        int idx = 0;
        for (int j = 1;j <= s;j <<= 1) ww[++idx] = j * w[num[i]],vv[idx] = j * v[num[i]],s -= j;
        if (s) ww[++idx] = s * w[num[i]],vv[idx] = s * v[num[i]];
        for (int j = m;j >= w[num[i]];j--) tomax (f[i][j],f[i + 1][j - w[num[i]]] + v[num[i]]);
        for (int k = 1;k <= idx;k++) {
            for (int j = m;j >= ww[k];j--) tomax (f[i][j],f[i][j - ww[k]] + vv[k]);
        }
        if (i > 1) {
            for (int j = 0;j <= m;j++) tomax (f[i][j],f[last[num[i]] + 1][j]);
        }
    }
    tomax (ans,f[1][m]);
    for (int i = 1;i <= timestamp;i++) for (int j = 0;j <= m;j++) f[i][j] = 0;
}
void solve (int _u) {
    get_root (_u,-1,get_size (_u,-1));
    int u = rt;
    calc (u);
    del[u] = 1;
    for (int v : g[u]) {
        if (del[v]) continue;
        solve (v);
    }
}
void mian () {
    cin >> n >> m;
    for (int i = 1;i <= n;i++) g[i].clear (),del[i] = 0;
    for (int i = 1;i <= n;i++) cin >> v[i];
    for (int i = 1;i <= n;i++) cin >> w[i];
    for (int i = 1;i <= n;i++) cin >> d[i];
    for (int i = 1;i <= n - 1;i++) {
        int a,b;
        cin >> a >> b;
        g[a].pb (b),g[b].pb (a);
    }
    ans = 0;
    solve (1);
    cout << ans << endl;
}
int main () {
	int T = 1;
	cin >> T;
	while (T--) mian ();
	return 0;
}