Overview

Given the initial states of \(N\) dice, this problem asks us to simulate \(Q\) operations (rolling a specific die 1 square to the left or right) and find the final position and top face value of each die.

Analysis

Since the number of squares \(M\) is as large as \(10^9\), an approach like “storing the entire grid in an array” will exceed the memory and time limits.

However, by noticing the following points, we can solve the problem very simply: 1. The dice are independent of each other: The movement of one die does not affect any other dice. 2. Very little information needs to be tracked: For each die, we only need to maintain three pieces of information: the “current position”, the “top face value”, and the “front face value”. 3. The rolling operation can be computed in \(O(1)\): When a die is rolled by 1 square, the new top and front face values can be immediately determined from the top and front face values before rolling.

The number of operations \(Q\) is at most \(2 \times 10^5\). If we can process each roll operation in \(O(1)\) time, simulating all \(Q\) operations will easily run within the time limit.

Algorithm

We manage the information for each die using a structure or class, and update the state of the die for each query.

For each die \(i\), we store the following information: - pos: Current position - u: Current top face value - f: Current front face value

When an operation \((d, c)\) is given, letting \(u\) and \(f\) be the top and front face values of die \(d\) before the update, respectively, we update the state according to direction \(c\) as follows:

• When c == '+' (direction where the square index increases):

  • Position: \(\text{pos} \leftarrow \text{pos} + 1\)
  • New top face: \(7 - f\)
  • New front face: \(u\)

• When c == '-' (direction where the square index decreases):

  • Position: \(\text{pos} \leftarrow \text{pos} - 1\)
  • New top face: \(f\)
  • New front face: \(7 - u\)

After simulating all operations, output the final position pos and top face u for each die.

Complexity

• Time Complexity: \(O(N + Q)\)
  • Reading the initial states of the dice takes \(O(N)\).
  • Processing each query (simulation) takes \(O(1) \times Q = O(Q)\).
  • Outputting the final results takes \(O(N)\).
  • In total, it is \(O(N + Q)\), which easily fits within the time limit.
• Space Complexity: \(O(N)\)
  • \(O(N)\) memory is used to store the states (position, top face, front face) of the \(N\) dice.

Implementation Notes

• Saving values before updating: When updating the state, computing the new top face requires the front face value prior to the update, and computing the new front face requires the top face value prior to the update. Make sure to store the original \(u\) and \(f\) values in temporary variables before performing the update.

• Choice of data types: Since the square position \(M\) can be up to \(10^9\), using a 64-bit integer type (such as long long in C++) for variables representing positions will safely prevent overflow.

  Source Code

#include <iostream>
#include <vector>

using namespace std;

struct Dice {
    long long pos;
    int u;
    int f;
};

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

    int N, Q;
    long long M;
    if (!(cin >> N >> M >> Q)) return 0;

    vector<Dice> dice(N + 1);
    for (int i = 1; i <= N; ++i) {
        cin >> dice[i].pos >> dice[i].u >> dice[i].f;
    }

    for (int j = 0; j < Q; ++j) {
        int d;
        char c;
        cin >> d >> c;

        int cur_u = dice[d].u;
        int cur_f = dice[d].f;

        if (c == '+') {
            dice[d].pos += 1;
            dice[d].u = 7 - cur_f;
            dice[d].f = cur_u;
        } else {
            dice[d].pos -= 1;
            dice[d].u = cur_f;
            dice[d].f = 7 - cur_u;
        }
    }

    for (int i = 1; i <= N; ++i) {
        cout << dice[i].pos << " " << dice[i].u << "\n";
    }

    return 0;
}

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