对上面有要求的树形dp：0115A

本文搬运自本人高中时期CSDN博客，若图片加载不出来，可到原文查看：https://blog.csdn.net/zhangtingxiqwq/article/details/135608809

http://47.92.197.167:5283/contest/451/problem/1

我初始的思路是维护一个 dp(i,j)dp(i,j) ，表示以 ii 根，向下染黑最远 jj 层，但这样子统计贡献就很难维护。

不妨换个思路，这个 jj 变成往上钦定要至少 jj 层。那样子每个染黑的时候的贡献是已经确定了的。

而对于儿子的 jj ，往上转移时只会转移到 j−1,j,j+1j-1,j,j+1

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// ubsan: undefined
// accoders
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
#define debug(...) void(0)
#endif
#define int long long
inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
#define Z(x) (x) * (x)
#define pb push_back
#define fi first
#define se second
//#define M
//#define mo
#define N 5010
int n, m, i, j, k, T;
int w[N], val[N];

namespace Sol1 {
int cr[N], b[N], ans, u, v, l, r;
vector<int> G[N];
//	queue<int> q;
int q[N];
int check() {
    int i, sum = 0, l, r;
    l = r = 0;
    //		memset(cr, -1, sizeof(cr));
    for (i = 1; i <= n; ++i) cr[i] = -1;
    for (i = 1; i <= n; ++i)
        if (!b[i])
            cr[i] = 0, q[++r] = i;
    if (l == r)
        return val[n - 1] * n;
    while (l < r) {
        u = q[++l];
        for (int v : G[u])
            if (cr[v] == -1)
                cr[v] = cr[u] + 1, q[++r] = v;
    }
    for (i = 1; i <= n; ++i)
        if (cr[i] > 0)
            sum += val[cr[i] - 1];
    //		debug(">> %lld \n", sum);
    return sum;
}
void dfs(int x, int s) {
    if (x > n) {
        ans = max(ans, check() - s);
        return;
    }
    b[x] = 0;
    dfs(x + 1, s);
    b[x] = 1;
    dfs(x + 1, s + w[x]);
}
void Main() {
    ans = 0;
    for (i = 1; i < n; ++i) {
        u = read();
        v = read();
        G[u].pb(v);
        G[v].pb(u);
    }
    dfs(1, 0);
    printf("%lld", ans);
}
}  // namespace Sol1

namespace Sol3 {
int dp[N][N], ans, u, v, sum = 0;
vector<int> G[N];
void Mx(int &a, int b) { a = max(a, b); }
void dfs(int x, int fa) {
    ++sum;
    int flg = 0;
    dp[x][0] = 0;
    for (int j = 1; j <= n; ++j) dp[x][j] = val[j - 1] - w[x];
    for (int y : G[x])
        if (y != fa) {
            dfs(y, x);
            for (int j = 0; j <= n; ++j) {
                k = dp[y][j];
                k = max(k, dp[y][j + 1]);
                if (j)
                    k = max(k, dp[y][j - 1]);
                dp[x][j] += k;
            }
        }
}
void Main() {
    //		for(i = 1; i <= n; ++i) ans += val[n - 1] - w[i];
    memset(dp, 0x80, sizeof(dp));
    for (i = 1; i < n; ++i) {
        u = read();
        v = read();
        G[u].pb(v);
        G[v].pb(u);
    }
    dfs(1, 0);
    //		assert(sum == n);
    for (i = 0; i <= n; ++i) {
        //			debug("%lld : %lld\n", i, dp[1][i]);
        Mx(ans, dp[1][i]);
    }
    printf("%lld", ans);
}
}  // namespace Sol3

signed main() {
    freopen("duck.in", "r", stdin);
    freopen("duck.out", "w", stdout);
    //	srand(time(NULL));
    //	T=read();
    //	while(T--) {
    //
    //	}
    n = read();
    for (i = 1; i <= n; ++i) w[i] = read();
    for (i = 0; i < n; ++i) val[i] = read();
    //	if(n <= 20) return Sol1 :: Main(), 0;
    Sol3 ::Main();
    //	Sol2 :: Main();
    return 0;
}