算法1

(暴力枚举) $O(n^2)$

初始矩阵各项均为0, 每读入一个地雷, 就把该地雷附近8个格子的值加1

C++ 代码

#include <iostream>
using namespace std;

const int N = 110;
int a[N][N];
bool mine[N][N];

void update(int x, int y) {
    int dx[] = {-1, -1, -1, 0, 0, +1, +1, +1}, dy[] = {-1, 0, +1, -1, +1, -1, 0, +1};
    for (int i = 0; i<8; i++) {
        a[x+dx[i]][y+dy[i]]++;
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    for (int i=1; i<=n; i++) {
        for (int j=1; j<=m; j++) {
            char c;
            cin >> c;
            if (c == '*') {
                mine[i][j] = true;
                update(i, j);
            }
        }
    }

    for (int i=1; i<=n; i++) {
        for (int j=1; j<=m; j++) {
            if (mine[i][j]) cout << '*';
            else cout << a[i][j];
        }
        cout << endl;
    }

    return 0;
}