在9*9的矩阵中随机生成10个“雷”，输出“雷”的分布矩阵（扫雷游戏的初始状态）。要求“雷”的位置输出字符（ASCII码15），非“雷”的位置输出附近有多少颗“雷”，附近无雷不输出。
提示：
① 声明一个9×9的二维数组，并将每个元素初始化为0；
② 随机生成10个“雷”：可使用随机函数生成10对不重复的行列下标，将其值置为-1，之后使用循环累计每个非-1元素周围-1的个数。
③ 计算非“雷”位置附近有多少“雷”，即当数组元素不等于-1时，计算它上下左右8个数组元素中值为-1的个数。可以累加这几个数组元素是否为-1的关系表达式和逻辑表达式之和来完成。例如计算数组元素a[i][j]附近上一行的“雷”个数可以使用如下表达式：
“(i>0&&j>0&&a[i-1][j-1]==-1)+(i>0&&a[i-1][j]==-1)+(i>0&&j<8&&a[i-1][j+1]==-1)”。

方法1

#include <iostream>
#include <random>
#include <time.h>
void main()
{
    char matrix[9][9];
    for (int i = 0; i < 9; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            matrix[i][j] = 0;
        }
    }

    srand(time(NULL));

    int n = 10;
    while (n--)
    {
        int line = rand() % 9;
        int row = rand() % 9;
        if (matrix[line][row] == 0)
        {
            matrix[line][row] = -1;
        }
        else
        {
            n++;
            continue;
        }
    }

    for (int i = 0; i < 9; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            if (matrix[i][j] != -1)
            {
                int leftIndex = j - 1;
                int rightIndex = j + 1;
                int topIndex = i - 1;
                int bottomIndex = i + 1;

                int left = leftIndex;
                int top = topIndex;

                int leftMoveStep = 0;
                int topMoveStep = 0;
                int count = 0;
                while (true)
                {
                    if (left >= 0 && left < 9 && top >= 0 && top < 9)
                    {
                        if (matrix[top][left] == -1)
                        { 
                            count++;
                        }
                    }

                    if (left == leftIndex && top == i)
                    {
                        break;
                    }

                    if (left - leftIndex < 2 && top == topIndex)
                    {
                        leftMoveStep = 1;
                        topMoveStep = 0;
                    }
                    else if (left == rightIndex && top == topIndex)
                    {
                        topMoveStep = 1;
                        leftMoveStep = 0;
                    }
                    else if (left == rightIndex && top == bottomIndex)
                    {
                        topMoveStep = 0;
                        leftMoveStep = -1;
                    }
                    else if (left == leftIndex && top == bottomIndex)
                    {
                        leftMoveStep = 0;
                        topMoveStep = -1;
                    }
                    else if (left - leftIndex < 2 && top == bottomIndex)
                    {
                        leftMoveStep = -1;
                        topMoveStep = 0;
                    }
                    left += leftMoveStep;
                    top += topMoveStep;
                }

                matrix[i][j] = count;

                count > 0 ? printf("%d ", count) : printf("  ");
            }
            else
            {
                printf("%c ", 15);
            }
        }
        printf("\n");
    }

    getchar();
}

方法2(性能更好)

#include <iostream>
#include <random>
#include <time.h>
void main()
{
    char matrix[9][9];
    for (int i = 0; i < 9; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            matrix[i][j] = 0;
        }
    }

    srand(time(NULL));

    int n = 10;
    while (n--)
    {
        int line = rand() % 9;
        int row = rand() % 9;
        if (matrix[line][row] != -1)
        {
            matrix[line][row] = -1;

            int leftIndex = row - 1;
            int rightIndex = row + 1;
            int topIndex = line - 1;
            int bottomIndex = line + 1;

            int left = leftIndex;
            int top = topIndex;

            int leftMoveStep = 0;
            int topMoveStep = 0;
            int count = 0;
            while (true)
            {
                if (left >= 0 && left < 9 && top >= 0 && top < 9)
                {
                    if (matrix[top][left] != -1)
                    {
                        matrix[top][left] += 1;
                    }
                }

                if (left == leftIndex && top == line)
                {
                    break;
                }

                if (left - leftIndex < 2 && top == topIndex)
                {
                    leftMoveStep = 1;
                    topMoveStep = 0;
                }
                else if (left == rightIndex && top == topIndex)
                {
                    topMoveStep = 1;
                    leftMoveStep = 0;
                }
                else if (left == rightIndex && top == bottomIndex)
                {
                    topMoveStep = 0;
                    leftMoveStep = -1;
                }
                else if (left == leftIndex && top == bottomIndex)
                {
                    leftMoveStep = 0;
                    topMoveStep = -1;
                }
                else if (left - leftIndex < 2 && top == bottomIndex)
                {
                    leftMoveStep = -1;
                    topMoveStep = 0;
                }
                left += leftMoveStep;
                top += topMoveStep;
            }
        }
        else
        {
            n++;
            continue;
        }
    }

    for (int i = 0; i < 9; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            if (matrix[i][j] != -1)
            {

                matrix[i][j] > 0 ? printf("%d ", matrix[i][j]) : printf("  ");
            }
            else
            {
                printf("%c ", 15);
            }
        }
        printf("\n");
    }

    getchar();
}

https://blog.csdn.net/vicjoe_hwakoo/article/details/7439307