原题链接：http://www.dotcpp.com/oj/problem1096.html

题目描述

Minesweeper Have you ever played Minesweeper? This cute little game comes with a certain operating system whose name we can’t remember. The goal of the game is to find where all the mines are located within a M x N field. The game shows a number in a square which tells you how many mines there are adjacent to that square. Each square has at most eight adjacent squares. The 4 x 4 field on the left contains two mines, each represented by a ``*’’ character. If we represent the same field by the hint numbers described above, we end up with the field on the right:
+。。。
。。。。
。+。。
。。。。

+100
2210
1+10
1110
（此处用”+“代替“*”用“。”代替“.”）

输入

The input will consist of an arbitrary number of fields. The first line of each field contains two integers n and m ( 0 < n, m <100) which stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly m characters, representing the field. Safe squares are denoted by .'' and mine squares by*,’’ both without the quotes. The first field line where n = m = 0 represents the end of input and should not be processed.

输出

For each field, print the message Field #x: on a line alone, where x stands for the number of the field starting from 1. The next n lines should contain the field with the ``.’’ characters replaced by the number of mines adjacent to that square. There must be an empty line between field outputs.

样例输入

4 4
+。。。
。。。。
。+。。
。。。。
3 5
++。。。
。。。。。
。+。。。
0 0

样例输出

Field #1:
+100
2210
1+10
1110

Field #2:
++100
33200
1+100

解题思路

这里注意一下，由于编辑器的问题，我把星号和点用加号和句号替代了，希望能理解，也能看一下链接的，那是原题（未改动的）

• 这里主要是英文的理解问题，可以通过使用翻译工具（针对英语菜鸡，比如我）大致理解一下题意：
• 你应该有玩过windows里的一个小游戏叫做"挖地雷"。这个游戏的目的就是要在M*N的地雷区格子中找出所有的地雷。
  为了要帮助你，这个游戏会在非地雷的格子上有些数字，告诉你这一个格子的邻居共有多少个地雷。
• 所以这道题就是通过输入的点 输出每个点周围有多少个地雷（是地雷就原样输出地雷），每一个点，除了周围的，均有8个邻居，你就需要判断这八个邻居有多少个是地雷，周围的可以使用条件判断一下，也就可以了，我在这里主要是：
• 通过给定的点，给予给定坐标在-1到1之间进行加减，从而达到遍历周围的八个点，那么具体看代码吧

参考代码（c++描述）

#include<iostream>
#include<cstdio>
using  namespace std;
char a[101][101];
int c(int dx, int dy, char a[][101]);
int n, m;
int main()
{

	cin >> n >> m;
	getchar();
	int t = 0, k,b = 1;
	while (n != 0 || m != 0)
	{
		t = 0;
		//输入操作 
		for (int i = 0; i < n; ++i)
		{
			for (int j = 0; j < m; ++j)
			{
				cin >> a[i][j];
			}
		}
		
		cout << "Field #" << b << ":" << endl;
		for (int i = 0; i < n; ++i)
		{
			for (int j = 0; j < m; ++j)
			{
				if (a[i][j] == '*') 
				{
					cout << '*';
				}
				else
				{
					k = c(i, j, a);
					cout << k;
				}
				++t;
				if (t % m == 0)
				{
					cout << endl;
				}


			}
		}
		cout << endl;
		cin >> n >> m;
		getchar();
		++b;
	}
	return 0;
}

int c(int dx, int dy, char a[][101])
{
	int count = 0;
	int p = 0, q = 0;
	for (int i = -1; i <= 1; ++i)
	{
		for (int j = -1; j <= 1; ++j)
		{
			p = dx + i;
			q = dy + j;
			if (p < 0 || p >= n || q < 0 || q >= m)
			{
				continue;
			}
			else
			{
				if (a[p][q] == '*')
				{
					++count;
				}

			}


		}
	}
	return count;
}

核心算法：https://blog.csdn.net/weixin_42792088/article/details/87907514