扫雷程序超时

Question

我正在学习普林斯顿大学在 cousera 上的课程，我针对以下问题编写了以下代码：

Minesweeper. Minesweeper is a 1960s era video game played on an m-by-n grid of cells. The goal is to deduce which cells contain hidden mines using clues about the number of mines in neighboring cells. Write a program Minesweeper.java that takes three integer command-line arguments m, n, and k and prints an m-by-n grid of cells with k mines, using asterisks for mines and integers for the neighboring mine counts (with two space characters between each cell). To do so, Generate an m-by-n grid of cells, with exactly k of the mn cells containing mines, uniformly at random. For each cell not containing a mine, count the number of neighboring mines (above, below, left, right, or diagonal).

每次提交给评分员；它说超过了 60 秒的限制，并停止评分。

这是期望的输出

public class Minesweeper {

    public static void main(String[] args) {

        int m = Integer.parseInt(args[0]);
        int n = Integer.parseInt(args[1]);
        int k = Integer.parseInt(args[2]);

        boolean[][] minePositions = new boolean[m + 2][n + 2];
        int[][] grid = new int[m + 2][n + 2];

        int num = 0;
        while (num != k) {
            if ((m * n) == k) {
                for (int f = 1; f <= m; f++) {
                    for (int z = 1; z <= n; z++) {
                        minePositions[f][z] = true;
                    }
                }
                break;
            }
            int r = (int) (Math.random() * (m * n - 1));
            int q = r / n;
            int rem = r % n;
            if (q == 0) {
                q = 1;
            }
            else if (q > (m - 1)) {
                q = m - 1;
            }
            if (rem == 0) {
                rem = 1;
            }
            else if (rem > (n - 1)) {
                rem = n - 1;
            }
            
            if (!minePositions[q][rem]) {
                minePositions[q][rem] = true;
                num++;
            }
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (minePositions[i - 1][j + 1]) grid[i][j]++;
                if (minePositions[i][j + 1]) grid[i][j]++;
                if (minePositions[i + 1][j + 1]) grid[i][j]++;
                if (minePositions[i - 1][j]) grid[i][j]++;
                if (minePositions[i + 1][j]) grid[i][j]++;
                if (minePositions[i - 1][j - 1]) grid[i][j]++;
                if (minePositions[i][j - 1]) grid[i][j]++;
                if (minePositions[i + 1][j - 1]) grid[i][j]++;
            }
            for (int j = 1; j <= n; j++) {
                if (minePositions[i][j]) System.out.print("*  ");
                else System.out.print(grid[i][j] + "  ");
            }
            System.out.print("\n");
        }

    }
}

Answer 1

我认为使用大小为 (m + 2, n + 2) 的数组会混淆您的数学运算。我会继续你的工作，在你的计算中总是使用 m + 2 和 n + 2 以避免任何混淆。

int r = (int) (Math.random() * ((m + 2) * (n + 2)));

生成一个可以跨越所有数组单元格的随机数。

int q = r / (n + 2);
int rem = r % (n + 2);

// Increment num only if q and rem are in the ranges and the position does not have a mine on it
if (q > 0 && q <= m && rem > 0 && rem <= n && !minePositions[q][rem]) {
    minePositions[q][rem] = true;
    num++;
}

完整代码

int m = Integer.parseInt(args[0]);
int n = Integer.parseInt(args[1]);
int k = Integer.parseInt(args[2]);

boolean[][] minePositions = new boolean[m + 2][n + 2];
int[][] grid = new int[m + 2][n + 2];

int num = 0;
while (num != k) {
    int r = (int) (Math.random() * ((m + 2) * (n + 2)));
    int q = r / (n + 2);
    int rem = r % (n + 2);

    if (q > 0 && q <= m && rem > 0 && rem <= n && !minePositions[q][rem]) {
        minePositions[q][rem] = true;
        num++;
    }
}

for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
        if (minePositions[i - 1][j + 1]) grid[i][j]++;
        if (minePositions[i][j + 1]) grid[i][j]++;
        if (minePositions[i + 1][j + 1]) grid[i][j]++;
        if (minePositions[i - 1][j]) grid[i][j]++;
        if (minePositions[i + 1][j]) grid[i][j]++;
        if (minePositions[i - 1][j - 1]) grid[i][j]++;
        if (minePositions[i][j - 1]) grid[i][j]++;
        if (minePositions[i + 1][j - 1]) grid[i][j]++;
    }
    for (int j = 1; j <= n; j++) {
        if (minePositions[i][j]) System.out.print("*  ");
        else System.out.print(grid[i][j] + "  ");
    }
    System.out.print("\n");
}

扫雷程序超时

Minesweeper program exceeding time limit

java

optimization

time

minesweeper