Overview

This problem asks you to arrange \(N\) signboards (strings) in order, maximize the overlap between adjacent signboards when concatenating them, and find the total number of characters in the resulting banner. There is a constraint on the overlap length: “neither string is completely contained within the overlap (each string must have at least \(1\) character remaining outside the overlap).”

Analysis

Naive Approach and Its Limitations

Consider finding the overlap length \(l\) for two adjacent strings \(S_{i-1}\) and \(S_i\). If we naively compare characters to check whether the suffix and prefix match for every \(l\) (\(1 \leq l < \min(|S_{i-1}|, |S_i|)\)), it takes \(O(|S_{i-1}| \times |S_i|)\) time in the worst case for a single pair. Given the constraint that the total length of all strings \(\sum |S_i| \leq 10^6\), this method will not meet the time limit (TLE).

Therefore, we need a method to efficiently find the longest length where “the suffix of one string” matches “the prefix of another string”.

Application of KMP (Prefix Function)

As an algorithm for efficiently determining string matches, the Prefix function (\(\pi\) array) from the KMP algorithm (Knuth-Morris-Pratt Algorithm) can be used.

The Prefix function \(\pi[j]\) for a string \(T\) represents “the length of the longest proper prefix of the substring consisting of the first \(j+1\) characters of \(T\) that is also a suffix.”

To utilize this, we construct a new string \(T\) as follows: $\(T = S_i + \text{"\#"} + S_{i-1}\)\( Here, `#` is a special character that does not appear in \)Si\( or \)S{i-1}$.

When we compute the Prefix function for this string \(T\), the value at the last element \(\pi[|T|-1]\) directly represents the longest length that is both a “prefix of \(S_i\)” and a “suffix of \(S_{i-1}\)”. Since the special character # is inserted between them, the match length cannot exceed the length of \(S_i\).

Handling the “Not Completely Contained in the Overlap” Condition

The problem has an important constraint: “\(l < |S_{i-1}|\) and \(l < |S_i|\)”. If the longest match length \(v = \pi[|T|-1]\) obtained from the Prefix function does not satisfy this condition (i.e., \(v \geq |S_{i-1}|\) or \(v \geq |S_i|\)), then that overlap is not allowed.

In this case, we use a property of the KMP algorithm to fall back to the next longest match candidate. Specifically, we repeatedly apply the update \(v = \pi[v-1]\) until the condition is satisfied. This allows us to efficiently find the “longest match length” within the valid range.

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Algorithm

For each adjacent pair of signboards \((S_{i-1}, S_i)\), perform the following steps:

1. Create a temporary string: Create \(T = S_i + \text{"\#"} + S_{i-1}\).
2. Compute the Prefix function: Compute the \(\pi\) array of the KMP algorithm for string \(T\).
3. Search for the maximum length satisfying the condition: Set \(v = \pi[|T|-1]\). While \(v \geq |S_{i-1}|\) or \(v \geq |S_i|\), keep updating \(v = \pi[v-1]\).
4. Update the total character count: The \(v\) that satisfies the condition becomes the overlap length \(L_i\). Add \(|S_i| - L_i\) to the total character count.

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Complexity

Time Complexity: \(O(\sum_{i=1}^{N} |S_i|)\)

The computation of the KMP Prefix function runs in time proportional to the length of string \(T\), i.e., \(O(|T|) = O(|S_{i-1}| + |S_i|)\). Additionally, the process of traversing back through the \(\pi\) array until the condition is met (v = pi[v-1]) is bounded by linear time overall, since \(v\) decreases at each step. Even when combining the processing for concatenating all signboards, the overall complexity is \(O(\sum |S_i|)\), proportional to the total length of all strings, which is well within the time limit.

Space Complexity: \(O(\sum_{i=1}^{N} |S_i|)\)

We need memory for the array holding all strings and the \(\pi\) array. Since these are at most proportional to the total length of the input strings, the space complexity is \(O(\sum |S_i|)\), which comfortably fits within the memory limit.

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Implementation Notes

• Fast I/O: In C++, since the input data is large, we use cin.tie(NULL) and ios_base::sync_with_stdio(false) to speed up standard I/O.

• Memory Reuse: Allocating a new \(\pi\) array at every step would incur memory allocation overhead, so we reuse a vector<int> pi and resize it as needed for better performance.

• Optimizing String Concatenation: When concatenating strings like t += s[i], we use reserve beforehand to allocate the necessary memory space, preventing unnecessary copies caused by dynamic memory reallocation.

  Source Code

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

using namespace std;

int main() {
    // Optimize standard I/O operations for competitive programming
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

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

    vector<string> s(n);
    for (int i = 0; i < n; i++) {
        cin >> s[i];
    }

    long long total_len = s[0].length();
    vector<int> pi;

    for (int i = 1; i < n; i++) {
        // Construct the string: S_i + "#" + S_{i-1}
        string t;
        t.reserve(s[i].length() + 1 + s[i-1].length());
        t += s[i];
        t += '#';
        t += s[i-1];

        int m = t.length();
        if (pi.size() < m) {
            pi.resize(m);
        }
        
        // Compute the KMP prefix function
        pi[0] = 0;
        for (int j = 1; j < m; j++) {
            int k = pi[j-1];
            while (k > 0 && t[j] != t[k]) {
                k = pi[k-1];
            }
            if (t[j] == t[k]) {
                k++;
            }
            pi[j] = k;
        }

        int v = pi[m-1];
        int len_prev = s[i-1].length();
        int len_curr = s[i].length();
        
        // Enforce the strict condition: overlap length L_i < |S_{i-1}| and L_i < |S_i|
        while (v > 0 && (v >= len_prev || v >= len_curr)) {
            v = pi[v-1];
        }
        
        total_len += (len_curr - v);
    }

    cout << total_len << "\n";

    return 0;
}

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