Kattis dancerecital solution

Dance Recital

In this article, we will dive into the provided solution for the Dance Recital problem from Kattis. This solution is written in Java and is engineered with a performance-first approach, which is crucial given the problem's complexity and input limitations.

Problem Overview

In this problem, each dancer is represented by a unique uppercase letter, and each dance routine is described by a string of characters, each character representing a dancer. The key challenge is to rearrange the order of these routines to minimize the number of quick costume changes needed when dancers perform in consecutive routines.

Solution Strategy

The solution employs an exhaustive search over all possible permutations of the routines. This brute-force approach ensures that no possible ordering is missed, crucial for finding the optimal solution. The hint from the CP-book suggests trying all R! permutations and comparing adjacent routines, which forms the backbone of our solution.

github full java solution

The Solution Explained

Reading the Input

The program begins by reading the number of routines R and then reads each routine, converting the characters into an integer bitmask. This bitmask is an efficient way to represent the dancers in each routine.

int R = Integer.parseInt(in.readLine());
int[] recitals = new int[R];
for (int i = 0; i < R; i++) {
    for (char c : in.readLine().toCharArray()) {
        recitals[i] |= 1 << (c - 'A');
    }
}

Calculating Distances Between Routines

Next, a 2D array distances is created to store the number of common dancers between each pair of routines. This step is crucial as it pre-calculates the quick changes required between any two routines, optimizing the performance of the algorithm.

int[][] distances = new int[R][R];
for (int i = 0; i < R; i++) {
    for (int j = i + 1; j < R; j++) {
        distances[i][j] = Integer.bitCount(recitals[i] & recitals[j]);
    }
}

Generating All Permutations

The core of the solution lies in generating all possible permutations of the routines. The indices array represents the current permutation of routines.

int[] indices = new int[R];
for (int i = 0; i < R; i++) {
    indices[i] = i;
}

Finding the Minimum Sum of Quick Changes

For each permutation, the program calculates the sum of quick changes required for that particular ordering. The result is updated whenever a permutation with a lower sum is found.

int permutations = factorial(R);
int result = Integer.MAX_VALUE;
for (int i = 0; i < permutations; i++) {
    int sum = 0;
    for (int j = 1; j < R; j++) {
        int minIndex = Math.min(indices[j - 1], indices[j]);
        int maxIndex = Math.max(indices[j - 1], indices[j]);
        sum += distances[minIndex][maxIndex];
    }
    if (sum < result) {
        result = sum;
    }
    nextPermutation(indices);
}

Outputting the Result

Finally, the minimum number of quick changes required is printed.

out.println(result);
out.close();

Conclusion

This solution to the "Dance Recital" problem exemplifies a brute-force approach tailored for competitive programming, where performance is paramount. By pre-calculating the quick change requirements between each pair of routines and exhaustively searching through all permutations, the solution ensures accuracy and efficiency, crucial for tackling such challenges.