We are given a collection of directories and files, and asked to find the directory from which the sum of relative paths to files is minimized. Abstractly, this is a graph problem: the collection of directories forms a tree, where the files are special leaves of the tree, and there are two weights on each edge. For instance, if the weight from node $u$ to node $v$ is $w_{uv}$, this means that any relative path ending at $u$ (but not passing through $v$) must be extended by $w$ characters to obtain a relative path for $v$. There is a corresponding weight $w_{vu}$, which may not be equal to $w_{uv}$.

So we want to find, for each node in the tree, the sum of path lengths from the node to special leaves. Our answer will essentially be the minimum sum, although depending on how exactly the weights are chosen, some care should be taken with counting backslash characters correctly.

Now the question is how to compute the sums. For each node, split the sum into an up-sum and a down-sum, where the down-sum counts paths to special leaves in the node's subtree, and the up-sum counts paths to all other special leaves. Then the down-sum of a node is almost equal to the sum over all its children of their down-sums: every special leaf in the node's subtree is counted by exactly one of the children. So we must simply account for how much each path increased. For a particular child, all paths in its subtree increased by the weight of the (directed) edge from the node to the child. This means that given the number of special leaves in each child's subtree, and the down-sum of each child, we can compute the down-sum of the parent node. Therefore we can compute all down-sums in a single depth-first search.

We can compute the up-sums in a second depth-first search, as follows. Fix some node $u$ with parent $p$. Every special leaf not in the subtree of $u$ is either not in the subtree of $p$, or in the subtree of a different child of $u$. So the up-sum of $u$ can be computed from the up-sum of $p$ (for the first case), the down-sum of $p$ minus the contribution of $u$ to this down-sum (for the second case), and the weight of the (directed) edge from $u$ to $p$ (since all paths increase by this length when extended from $p$ to $u$), and the number of special leaves not in the subtree of $p$ (since this is the number of paths whose lengths increase by that weight).

#include <cstdio>
#include <cassert>
#include <cstring>
#include <vector>
using namespace std;
 
#define NMAX 100000
 
struct Node {
  bool isFile;
  vector<Node*> children;
  int namelen;
 
  int numLeaves;
  long long totalSubtreeLen;
 
  long long total;
};
 
Node nodes[NMAX];
 
int n;
int nleaves;
 
void dfs1(Node* node) {
  node->numLeaves = (node->isFile ? 1 : 0);
  node->totalSubtreeLen = 0;
  for (Node* child : node->children) {
    dfs1(child);
    node->numLeaves += child->numLeaves;
    node->totalSubtreeLen += child->totalSubtreeLen + child->numLeaves * (child->namelen + (child->isFile ? 0 : 1));
  }
}
 
void dfs2(Node* node, long long parentlen) {
  node->total = parentlen + node->totalSubtreeLen;
 
  long long plenadd = 0;
  for (Node* child : node->children) {
    plenadd += child->totalSubtreeLen + child->numLeaves * (child->namelen + (child->isFile ? 0 : 1));
  }
  for (Node* child : node->children) {
    dfs2(child, parentlen + plenadd -
      (child->totalSubtreeLen + child->numLeaves * (child->namelen + (child->isFile ? 0 : 1)))
      + 3 * (nleaves - child->numLeaves));
  }
}
 
int main() {
  scanf("%d", &n);
  char name[40];
  nleaves = 0;
  for (int i = 0; i < n; i++) {
    scanf("%s", name);
    nodes[i].namelen = strlen(name);
    int numChildren;
    scanf("%d", &numChildren);
    nodes[i].isFile = (numChildren == 0);
    if (nodes[i].isFile) {
      nleaves++;
    }
    for (int j = 0; j < numChildren; j++) {
      int id;
      scanf("%d", &id);
      nodes[i].children.push_back(&nodes[id-1]);
    }
  }
 
  assert(!nodes[0].isFile);
 
  dfs1(&nodes[0]);
  dfs2(&nodes[0], 0);
  long long ans = nodes[0].total;
  for (int i = 0; i < n; i++) {
    if (!nodes[i].isFile) {
      ans = (ans < nodes[i].total ? ans : nodes[i].total);
    }
  }
  printf("%lld\n", ans);
}