Overview

This problem asks us to select one prefix for each of the given \(N\) strings such that no selected prefix is a prefix of another (i.e., they are “prefix-free”), while minimizing the total length of the selected prefixes.

Since this problem deals with prefixes of strings, it can be solved visually and efficiently using a Trie (prefix tree).

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Analysis

1. Reformulating the Problem (Interpretation on a Trie)

Consider inserting all given strings \(S_1, S_2, \ldots, S_N\) into a Trie. Each string \(S_i\) corresponds to a terminal node (target node) in the Trie.

Choosing a prefix \(T_i\) of each \(S_i\) is equivalent to “choosing one node on the path from the target node of \(S_i\) to the root”.

The prefix-free condition can be restated on the Trie as follows: * “The \(N\) chosen nodes must not have any ancestor-descendant relationship with each other.”

2. Which Node is Optimal to Choose?

Let \(u\) be the node chosen for string \(S_i\). Since we want to minimize the total length, we want to choose nodes as close to the root as possible (i.e., with small depth).

However, if we go too close to the root, it may interfere with other strings \(S_j\). Specifically, if the subtree of node \(u\) contains the target node of another string \(S_j\), then we cannot choose \(u\). This is because if we choose \(u\), the node chosen for \(S_j\) must necessarily be a descendant of \(u\), which would create an ancestor-descendant relationship (violating the prefix-free condition).

Therefore, the optimal node \(u\) to choose for each \(S_i\) is the shallowest (closest to the root) node satisfying the following: * “The subtree rooted at node \(u\) contains exactly \(1\) target node (only \(S_i\) itself).”

If no such node exists (excluding the root node representing the empty string), then a valid assignment is impossible. For example, if some string \(S_i\) is a prefix of another string \(S_j\), the subtree of \(S_i\)’s terminal node also contains \(S_j\)’s terminal node, making the target count \(2\) or more, and thus the assignment becomes impossible.

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Algorithm

Based on the above analysis, we can find the solution through the following steps.

Step 1: Build the Trie

Insert all \(N\) given strings into the Trie. Record the position of each string’s terminal node (target node).

Step 2: Count Targets in Each Subtree

For each node \(v\), compute “the number of target nodes \(C_v\) contained in the subtree of \(v\)”. This can be computed by traversing the Trie nodes from leaves (deeper nodes) toward the root (shallower nodes), accumulating values in a manner similar to dynamic programming on trees (tree DP).

Step 3: Determine the Optimal Prefix and Check Feasibility

For each string \(S_i\) (with target node \(u\)), perform the following search: 1. Traverse from \(u\) to its parent \(p\) one step at a time. 2. As long as the parent node \(p\) has subtree target count \(C_p\) equal to \(1\), move upward (to the parent). However, do not move to the root node \(0\) (which represents the empty string). 3. The node where you can no longer move upward is the optimal node to choose. 4. If the target count \(C\) of the final node reached is not \(1\), or if that node is the root node, then a valid prefix cannot be constructed, so output -1 and terminate. 5. Otherwise, add the depth of that node (= the length of the abbreviation) to the answer.

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Complexity

Let \(L = \sum_{i=1}^{N} |S_i|\) be the total length of all strings.

• Time Complexity: \(O(L)\)

  • Insertion into the Trie takes \(O(L)\).
  • Computing the subtree target count \(C\) requires a single reverse loop over the nodes (at most \(L+1\)), so it takes \(O(L)\).
  • The operation of traversing parents for each string passes through each node at most once, so it takes \(O(L)\) in total.
  • Therefore, the overall time complexity is \(O(L)\), which runs sufficiently fast for the constraint \(L \le 5 \times 10^5\).

• Space Complexity: \(O(L)\)

  • Since the number of Trie nodes is at most \(L+1\), the space complexity is also \(O(L)\).

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Implementation Notes

• Traversal in Topological Sort Order: When inserting strings into the Trie, newly created nodes are always appended to the nodes array after their parent nodes. Therefore, simply looping over node indices in reverse order (from V - 1 down to 1) naturally gives a traversal in topological sort order “from leaves to root”. This allows us to perform subtree aggregation simply without writing recursive functions (DFS).

  Source Code

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

struct Trie {
    struct Node {
        int next[26];
        int parent;
        int depth;
        bool is_target;
        int C;
        Node() {
            fill(next, next + 26, -1);
            parent = -1;
            depth = 0;
            is_target = false;
            C = 0;
        }
    };
    vector<Node> nodes;

    Trie() {
        nodes.emplace_back(); // 根ノード
    }

    int insert(const string& s) {
        int curr = 0;
        for (char c : s) {
            int idx = c - 'a';
            if (nodes[curr].next[idx] == -1) {
                int next_node = nodes.size();
                nodes.emplace_back();
                nodes[next_node].parent = curr;
                nodes[next_node].depth = nodes[curr].depth + 1;
                nodes[curr].next[idx] = next_node;
            }
            curr = nodes[curr].next[idx];
        }
        nodes[curr].is_target = true;
        return curr;
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N;
    if (!(cin >> N)) return 0;

    Trie trie;
    vector<int> targets(N);
    for (int i = 0; i < N; ++i) {
        string s;
        cin >> s;
        targets[i] = trie.insert(s);
    }

    int V = trie.nodes.size();
    for (int i = 0; i < V; ++i) {
        trie.nodes[i].C = trie.nodes[i].is_target ? 1 : 0;
    }

    for (int i = V - 1; i > 0; --i) {
        int p = trie.nodes[i].parent;
        trie.nodes[p].C += trie.nodes[i].C;
    }

    long long total_cost = 0;
    bool possible = true;

    for (int i = 0; i < N; ++i) {
        int u = targets[i];
        while (true) {
            int p = trie.nodes[u].parent;
            if (p == 0) {
                break;
            }
            if (trie.nodes[p].C != 1) {
                break;
            }
            u = p;
        }
        if (trie.nodes[u].C != 1 || u == 0) {
            possible = false;
            break;
        }
        total_cost += trie.nodes[u].depth;
    }

    if (possible) {
        cout << total_cost << "\n";
    } else {
        cout << -1 << "\n";
    }

    return 0;
}

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This editorial was generated by gemini-3.5-flash-thinking.