Uva Problem 10189 Minesweeper Problem

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 × 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 4×4 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
Input
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.
Output
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.
Sample Input
4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0
Sample Output
Field #1:
*100
2210
1*10
1110
Field #2:
**100
33200

1*100

解题方法转自：http://blog.youkuaiyun.com/metaphysis/article/details/6432031

// Minesweeper （扫雷）  
// PC/UVa IDs: 110102/10189, Popularity: A, Success rate: high Level: 1  
// Verdict: Accepted   
// Submission Date: 2011-04-09  
// UVa Run Time: 0.012s  
//  
// 版权所有（C）2011，邱秋。metaphysis # yeah dot net  
//  
// 解题思路很简单，使用字符数组表示给定的地雷图，逐个扫描统计各点周围的地雷总数。在计算地雷总数时，  
// 用到了一个小技巧，设置边界数组 bounds，将当前的点顺次加上边界数组的值即可得到当前点周围的 8 个  
// 点，注意不要忘记判断点是否在地雷地图范围内。  

#include <iostream>  
#include <cstring>  

using namespace std;  

#define MAXSIZE 100  

inline bool range_checking(int x, int y, int line, int row)  
{  
	return (0 <= x && x < line) && (0 <= y && y < row);  
}  

void display(char matrix[][MAXSIZE], int line, int row)  
{  
	int bounds[8][2] =  
	{ {-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0},  
	{1, 1} };  

	for (int i = 0; i < line; i++)  
	{  
		for (int j = 0; j < row; j++)  
		{  
			// 如果该位置为地雷则原样输出地雷的符号。  
			if (matrix[i][j] == '*')  
				cout << "*";  
			else  
			{  
				// 统计该点周围 8 点的地雷总数。  
				int mines = 0;  
				for (int k = 0; k < 8; k++)  
				{  
					/*
						将二维数组想象成为XY轴，
						先求X坐标,然后求Y坐标
					*/
					int m = i + bounds[k][0];  
					int n = j + bounds[k][1];  
					/*
						首先判断这个点是否在地雷地图范围内
						其次判断是否有雷
					*/
					if (range_checking(m, n, line, row)  
						&& matrix[m][n] == '*')  
						mines++;  
				}  

				cout << mines;  
			}  
		}  

		cout << endl;  
	}  
}  

int main(int ac, char *av[])  
{  
	char matrix[MAXSIZE][MAXSIZE];  
	int line, row, field = 0;  

	while ((cin >> line >> row, line && row))  
	{  
		memset(matrix, 0, sizeof(matrix));  

		for (int i = 0; i < line; i++)  
			for (int j = 0; j < row; j++)  
				cin >> matrix[i][j];  

		if (field > 0)  
			cout << endl;  
		field++;  

		cout << "Field #" << field << ":" << endl;  

		display(matrix, line, row);  
	}  

	return 0;  
}