思路
这道题看到所有c为0，那么所有树的生长时间就是确定的，那么就可以贪心的把生长时间长的先种上。
那么对于c不为0，我们考虑二分答案，二分完答案后，我们考虑对于每个树把他最晚种上的合法时间算出来，然后按照上面的贪心做就做完了。
不要写了一个函数不调用啊
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll a[100005], b[100005], c[100005];
inline __int128 sol(ll x, ll l, ll r)
{__int128 ret = (r - l + 1) * b[x];__int128 res = (r - l + 1);res *= l + r;res /= 2;res *= c[x];if (c[x] == 0){return ret;}if (c[x] > 0){return ret + res;}ll im = (1 - b[x]) / c[x];if (im < l){return r - l + 1;}if (im > r){return ret + res;}ret = (im - l + 1) * b[x];res = (im - l + 1);res *= l + im;res /= 2;res *= c[x];res += r - im;return ret + res;
}
int fa[100005];
vector<int> g[100005];
void dfs(ll x, ll f)
{fa[x] = f;for (auto v : g[x]){if (v == f){continue;}dfs(v, x);}
}
void write(__int128 x)
{if (x < 10){ll t = x % 10;cout << t;return;}ll t = x % 10;write(x / 10);cout << t;
}
int n;
int t[100005], p[100005];
bool cmp(int x, int y)
{return t[x] < t[y];
}
int dis[100005];
inline bool chk(ll x)
{// cout << x << "\n";for (int i = 1; i <= n; i++){if (sol(i, 1, x) < a[i]){return 0;}dis[i] = -1;p[i] = i;ll l = 1, r = n, rt = 0;while (l <= r){ll mid = (l + r) >> 1;if (sol(i, mid, x) >= a[i]){rt = mid;l = mid + 1;}else{r = mid - 1;}}t[i] = rt;// cout << t[i] << " \n"[i == n];}// for (int i = 1; i <= n; i++)// {// write(sol(i, t[i], x));// cout << " \n"[i == n];// }sort(p + 1, p + n + 1, cmp);dis[0] = 0;ll ntm = 0;for (int i = 1; i <= n; i++){ll nid = p[i];vector<int> v;while (dis[nid] == -1){v.push_back(nid);nid = fa[nid];}reverse(v.begin(), v.end());for (auto j : v){// cout << j << " " << ntm + 1 << "\n";dis[j] = ++ntm;}}for (int i = 1; i <= n; i++){// cout << dis[i] << " \n"[i == n];if (dis[i] > t[i]){return 0;}}return 1;
}
int main()
{ios::sync_with_stdio(0), cin.tie(0);cin >> n;for (int i = 1; i <= n; i++){cin >> a[i] >> b[i] >> c[i];}for (int i = 1; i < n; i++){int u, v;cin >> u >> v;g[u].push_back(v);g[v].push_back(u);}dfs(1, 0);int l = 0, r = 1e9;int ans = -1;while (l <= r){int mid = (l + r) >> 1;if (chk(mid)){r = mid - 1;ans = mid;}else{l = mid + 1;}}cout << ans << "\n";return 0;
}
 安装docker 28)
