原题传送门 如果以生产者开始为根建树的话 一个点可以向该点所有食物的 l c a lca lca以及这个 l c a lca lca的所有祖先贡献1灾难值 根据上面两句话，就可以把这道题给做掉了
不过还是讲一讲具体写法 先拓扑排序调整顺序（从高级消费者到低级生产者），然后从生产者开始建树，一边建一边初始化倍增数组 最终从最高级的消费者开始向下面的东西贡献灾难值
Code：
#include <bits/stdc++.h> #define maxn 1000010 using namespace std; struct Edge{ int to, next; }edge[maxn << 1]; int num, head[maxn], in[maxn], d[maxn], que[maxn], cnt, fa[maxn][25], size[maxn], n; queue <int> q; inline int read(){ int s = 0, w = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') w = -1; for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48); return s * w; } void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; } void topsort(){ for (int i = 1; i <= n; ++i) if (!in[i]) q.push(i); while (!q.empty()){ int u = q.front(); q.pop(); que[++cnt] = u; for (int i = head[u]; i; i = edge[i].next){ int v = edge[i].to; if (!(--in[v])) q.push(v); } } } int get(int u, int v){ if (d[u] > d[v]) swap(u, v); for (int i = 20; i >= 0; --i) if (d[u] <= d[fa[v][i]]) v = fa[v][i]; if (u == v) return u; for (int i = 20; i >= 0; --i) if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i]; return fa[u][0]; } int main(){ n = read(); for (int i = 1; i <= n; ++i){ int x = read(); while (x) ++in[x], addedge(i, x), x = read(); } topsort(); for (int i = n; i; --i){ int u = que[i], x = edge[head[u]].to; for (int j = edge[head[u]].next; j; j = edge[j].next) x = get(x, edge[j].to); fa[u][0] = x, d[u] = d[x] + 1; for (int j = 0; fa[u][j]; ++j) fa[u][j + 1] = fa[fa[u][j]][j]; } for (int i = 1; i <= n; ++i) ++size[que[i]], size[fa[que[i]][0]] += size[que[i]]; for (int i = 1; i <= n; ++i) printf("%d\n", size[i] - 1); return 0; }