思路

遍历一下地图，对所有的四周都无雷的格子进行bfs
做完后再遍历一下地图，补加上剩下的没有被bfs扫描到的格子（这些格子四周必有雷）

C++ 代码

#include<iostream>
#include<string.h>
#include<queue>
using  namespace std;

char map[305][305];
bool vis[305][305];
int dx[] = { -1,-1,-1,0,0,1,1,1 }, dy[] = { -1,0,1,-1,1,-1,0,1 };

void bfs(int x, int y, int n)
{
    queue<pair<int, int>> Q;
    Q.push({ x,y });
    vis[x][y] = 1;
    bool boom;
    int a, b;
    while (Q.size()) {
        auto top = Q.front();
        Q.pop();
        boom = 0;
        for (int k = 0; k < 8; k++) {
            a = top.first + dx[k], b = top.second + dy[k];
            if (a < 0 || a >= n || b < 0 || b >= n || vis[a][b]) continue;
            if (map[a][b] == '*') {
                boom = 1;
                break;
            }
        }
        if (!boom) {
            for (int i = 0; i < 8; i++) {
                a = top.first + dx[i], b = top.second + dy[i];
                if (a < 0 || a >= n || b < 0 || b >= n || vis[a][b]) continue;
                vis[a][b] = 1;
                Q.push({ a,b });
            }

        }
    }
}


int solve(int n)
{
    int a, b, res = 0;
    bool flag;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (map[i][j] == '*' || vis[i][j]) continue;
            flag = 0;
            for (int k = 0; k < 8; k++) {
                a = i + dx[k], b = j + dy[k];
                if (a < 0 || a >= n || b < 0 || b >= n || vis[a][b]) continue;
                if (map[a][b] == '*') {
                    flag = 1;
                    break;
                }
            }
            if (!flag) {
                bfs(i, j, n);   
                res++;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (map[i][j] == '.' && !vis[i][j]) res++;
        }
    }
    return res;
}


int main()
{
    int T, N;
    cin >> T;
    for (int i = 1; i <= T; i++) {
        cin >> N;
        memset(vis, 0, sizeof(vis));
        for (int j = 0; j < N; j++) cin >> map[j];
        cout << "Case #" << i << ": " << solve(N) << endl;
    }


    return 0;
}