对每个节点维护一个 $node$，$node_{ans}$ 是答案，$node_{ans1}$ 是 $\sum c_i\times(b_i+y)$，$node_{ans2}$ 是 $\sum c_i\times(a_i+x)$，$node_{sum}$ 是 $\sum c_i$，转移见代码。

// Expected: 100pts.

#include<bits/stdc++.h>
using namespace std;

const int maxN = 1e5 + 10;
using ll = long long;

int n, a[maxN], b[maxN], c[maxN], ch[maxN][2];

struct Node {
    ll ans, ans1, ans2, sum;
    inline Node operator + (const Node& rhs) const {
        Node res;
        res.ans = ans + rhs.ans; res.ans1 = ans1 + rhs.ans1;
        res.ans2 = ans2 + rhs.ans2; res.sum = sum + rhs.sum;
        return res;
    }
    inline bool operator < (const Node& rhs) const { return ans < rhs.ans; }
} nodes[maxN];

void dfs(int u) {
    if(u > n) {
        u -= n;
        return void(nodes[u + n] = {
                                    1ll * c[u] * a[u] * b[u], 
                                    1ll * c[u] * b[u],
                                    1ll * c[u] * a[u], 
                                    c[u]
                                });
    }
    dfs(ch[u][0]), dfs(ch[u][1]);
    // 翻新公路，右侧铁路不便利值增大
    Node rr = nodes[ch[u][1]]; rr.ans += rr.ans2; rr.ans1 += rr.sum;
    // 翻新铁路，左侧公路不便利值增大
    Node ll = nodes[ch[u][0]]; ll.ans += ll.ans1; ll.ans2 += ll.sum;
    nodes[u] = min(ll + nodes[ch[u][1]], rr + nodes[ch[u][0]]);
}

signed main() {
    cin >> n;
    for(int i = 1; i < n; i++) {
        cin >> ch[i][0] >> ch[i][1];
        if(ch[i][0] < 0) ch[i][0] = n - ch[i][0];
        if(ch[i][1] < 0) ch[i][1] = n - ch[i][1];
    }
    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];
    dfs(1);
    cout << nodes[1].ans;
}