Overview

This problem asks us to simulate rolling \(N\) dice \(Q\) times on a one-dimensional grid of squares, and output the final position and top face number of each die.

Analysis

The problem statement clearly defines the rules for how a die’s position and orientation (top face, front face) change when rolled. We can solve this problem by simply updating the state of the specified die according to these rules for each operation.

Looking at the constraints, the number of dice \(N\) and the number of operations \(Q\) are both up to \(2 \times 10^5\). Therefore, if we can process each operation in \(O(1)\) time, the total time complexity will be \(O(N+Q)\), which easily fits within the time limit. No complex data structures are needed, and a straightforward simulation is sufficient.

Algorithm

Prepare arrays to record the current position \(S\), top face number \(U\), and front face number \(F\) for each die. Read the \(Q\) operations one by one and update the state using the following steps:

Let the target die index be \(d\) and the direction be \(c\).

1. When \(c\) is + (direction of increasing square index):

   • Increase the position by \(1\) (\(S \leftarrow S + 1\)).
   • The new top face number becomes \(7 - F\), and the new front face number becomes the current top face number \(U\).

2. When \(c\) is - (direction of decreasing square index):

   • Decrease the position by \(1\) (\(S \leftarrow S - 1\)).
   • The new top face number becomes the current front face number \(F\), and the new front face number becomes \(7 - U\).

After all operations are completed, output the position \(S\) and top face \(U\) for each die in order. Note that, guaranteed by the problem constraints, a die will never move out of bounds, so there is no need to check whether a move is valid.

Complexity

• Time Complexity: \(O(N + Q)\)
  • It takes \(O(N)\) time to read the initial state of the dice and \(O(Q)\) time to process each operation.
• Space Complexity: \(O(N)\)
  • Required to store the state of each die in arrays.

Implementation Notes

• When updating the state of a die, the original front face is used to calculate the new top face, and the original top face is used to calculate the new front face. Therefore, you need to calculate both using the values prior to the update, or use temporary variables before assignment.

  
  // Example update in the '+' direction
  new_U = 7 - F;
  new_F = U;
  U = new_U;
  F = new_F;
  

• The die index \(d\) given in the input is 1-indexed (starts from 1), so do not forget to subtract 1 to adjust it to 0-indexed arrays.

• Although the maximum value of \(M\) is large (\(10^9\)), the actual number of moves is at most \(Q\). Therefore, an int type is sufficient for storing the position, but using a long long type is even safer.

  Source Code

#include <iostream>
#include <vector>

using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int N, M, Q;
    if (!(cin >> N >> M >> Q)) return 0;
    
    vector<long long> S(N), U(N), F(N);
    for (int i = 0; i < N; ++i) {
        cin >> S[i] >> U[i] >> F[i];
    }
    
    for (int i = 0; i < Q; ++i) {
        int d;
        char c;
        cin >> d >> c;
        --d;
        
        if (c == '+') {
            S[d]++;
            int new_U = 7 - F[d];
            int new_F = U[d];
            U[d] = new_U;
            F[d] = new_F;
        } else {
            S[d]--;
            int new_U = F[d];
            int new_F = 7 - U[d];
            U[d] = new_U;
            F[d] = new_F;
        }
    }
    
    for (int i = 0; i < N; ++i) {
        cout << S[i] << " " << U[i] << "\n";
    }
    
    return 0;
}

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