给定一颗有$n$个节点的树，点有点权，给定一个长度为$n$的排列$p$，给定一个长度为$n-1$的数组$c$，

我们会在树上进行$n-1$次行走，第$i$次我们会从$p[i]$走到$p[i+1]$，步长为$c[i]$，且走最短路，

当我们到$p[i+1]$的距离小于$c[i]$的时候，我们会直接到点$p[i+1]$，对每次行走，我们要求出其所经过点的点权和。

由于$n$只有$50000$，所以写一些比较暴力的算法，先选定根节点为$1$号节点：

如果$c[i]$大于$\sqrt{n}$，从$p[i]$到$p[i+1]$，最多只要跳$\sqrt{n}$个点，所以，可以直接暴力跳，然后找第$k$代祖先，单次操作$O(\sqrt{n}\log n)$的复杂度。

如果$c[i]$小于$\sqrt{n}$，我们定义$sum[rt][len]$，表示从点$rt$向上跳步长为$len$，跳出根节点的路径和，预处理这一步大概是$O(n\sqrt{n}\log \sqrt{n})$的复杂度。

之后只要求出$lca$，然后二分一下跳几次，在某条链上跳到哪，$k$代祖先找一下点，最后统计一下答案即可，整体复杂度$O(n\sqrt{n}\log n)$。

#include <bits/stdc++.h>using namespace std;const int N = 5e4 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;int sz[N], son[N], fa[N], id[N], rk[N], top[N], dep[N], tot;int a[N], p[N], c[N], n, block;long long s[N][250];void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs1(int rt, int f) {fa[rt] = f, dep[rt] = dep[f] + 1, sz[rt] = 1;for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs1(to[i], rt);sz[rt] += sz[to[i]];if (!son[rt] || sz[to[i]] > sz[son[rt]]) {son[rt] = to[i];}} 
}void dfs2(int rt, int tp) {top[rt] = tp, id[rt] = ++tot, rk[tot] = rt;if (!son[rt]) {return ;}dfs2(son[rt], tp);for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa[rt] || to[i] == son[rt]) {continue;}dfs2(to[i], to[i]);}
}int lca(int x, int y) {while (top[x] != top[y]) {if (dep[top[x]] < dep[top[y]]) {swap(x, y);}x = fa[top[x]];}return dep[x] < dep[y] ? x : y;
}int k_fa(int rt, int k) {if (dep[rt] <= k) {return 0;}while (k > dep[rt] - dep[top[rt]]) {k -= dep[rt] - dep[top[rt]] + 1;rt = fa[top[rt]];}return rk[id[rt] - k];
}void dfs3(int rt, int fa) {for (int i = 1; i <= block; i++) {s[rt][i] = a[rt] + s[k_fa(rt, i)][i];}for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa) {continue;}dfs3(to[i], rt);}
}long long f1(int u, int v, int c) {// 暴力跳int f = lca(u, v);if (dep[u] < dep[v]) {swap(u, v);}int cur = u;long long ans = a[u];while (dep[cur] - dep[f] >= c) {cur = k_fa(cur, c);ans += a[cur];}if (v == f) {// 只有一条链if (cur != v) {ans += a[v];}return ans;}if (dep[cur] + dep[v] - 2 * dep[f] <= c) {// 不能跳到链的另一端ans += a[v];return ans;}cur = k_fa(v, dep[v] - dep[f] - (c - (dep[cur] - dep[f])));ans += a[cur];while (dep[v] - dep[cur] >= c) {cur = k_fa(v, dep[v] - dep[cur] - c);ans += a[cur];}if (cur != v) {ans += a[v];}return ans;
}long long f2(int u, int v, int c) {// 优化跳。int f = lca(u, v);if (dep[u] < dep[v]) {swap(u, v);}long long ans = 0;int steps = (dep[u] - dep[f]) / c, cur = k_fa(u, steps * c), now;ans += s[u][c] - s[cur][c] + a[cur];if (v == f) { // 只有一条链if (cur != v) {ans += a[v];}return ans;}if (dep[cur] + dep[v] - 2 * dep[f] <= c) {// 不能跳到链的另一端ans += a[v];return ans;}cur = k_fa(v, dep[v] - dep[f] - (c - (dep[cur] - dep[f])));steps = (dep[v] - dep[cur]) / c, now = k_fa(v, dep[v] - dep[cur] - steps * c);ans += s[now][c] - s[cur][c] + a[cur], cur = now;if (cur != v) {ans += a[v];}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d", &n), block = sqrt(n);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 1, u, v; i < n; i++) {scanf("%d %d", &u, &v);add(u, v);add(v, u);}for (int i = 1; i <= n; i++) {scanf("%d", &p[i]);}for (int i = 1; i < n; i++) {scanf("%d", &c[i]);}dfs1(1, 0);dfs2(1, 1);dfs3(1, 0);for (int i = 1; i < n; i++) {printf("%lld\n", c[i] > block ? f1(p[i], p[i + 1], c[i]) : f2(p[i], p[i + 1], c[i]));}return 0;
}