P2670 [NOIP2015 普及组] 扫雷游戏

题目链接

https://www.luogu.com.cn/problem/P2670

题目思路

模拟当前点的八个方向,扫描是否有雷

题目代码

#include <iostream>
#include <algorithm>

using namespace std;
const int N = 110;
char g[N][N];
int n, m;

int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; i ++ ) cin >> g[i];
    
    int dx[8] = {-1, 0, 1, 0, -1, 1, -1, 1}, dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};
    for(int i = 0; i < n; i ++ )
    {
        for(int j = 0; j < m; j ++ )
        {
            int sum = 0;
            for(int k = 0; k < 8; k ++ )
            {
                if(g[i][j] == '*')
                {
                    cout << '*';
                    break;
                }
                int x = i + dx[k], y = j + dy[k];
                if(x >= 0 && x < n && y >= 0 && y < m && g[x][y] == '*') sum ++ ;
            }
            if(g[i][j] != '*') cout << sum;
        }
        puts("");
    }
    
    return 0;
}

标签：P2670,NOIP2015,题目,cout,int,sum,++,扫雷,&&
来源： https://www.cnblogs.com/vacilie/p/15983433.html