1419：Minesweeper

题目描述

ethan0ne has been particularly interested in minesweeper. It is a small game pre-installed on Windows 7 and before. Now ethan0ne has encountered several n × m minefields. You are asked to count the number of mines in adjacent cells of each cell in each matrix, with up to 8 adjacent cells per cell. 0 < n, m < = 100

输入

The input contains several matrices, and for each matrix, the first row contains two integers, n and m, representing the number of rows and columns of the matrix, respectively. The next n lines contain m characters each. Safe areas are denoted by ‘.’ and mine-prone areas by ‘*’. Input ends when n=m=0.

输出

For the i-th matrix, the sequence number is first printed on a single line: “Field # I:”, and on the next n lines, the ‘.’ read in should be replaced by the number of mines around that position. Each of the two matrices in the output must be separated by a blank line.

样例输入

4 4
…
…
.…
…
3 5
**…
…
.*…
0 0

样例输出

Field #1:
100
2210
110
1110
Field #2:
**100
33200
1*100

代码解答

#include <iostream>
using namespace std;

char trans(int num)
{
	for (int i = 0; i <= 9; i++)
	{
		if (num == i)
			return '0' + i;
	}
}

int main()
{
	char arr[100][100];
	int n, m; //代表行和列
	int i, j;
	int count = 0; //统计，方圆八里的雷有几个
	int field = 0;

	while (cin >> n >> m, m || n)
	{
		field++;

		//棋盘的输入
		for (i = 0; i < n; i++)
			cin >> arr[i];

		//统计，每一个格的数字应该是几
		for (i = 0; i < n; i++)
		{
			for (j = 0; j < m; j++)
			{
				if (arr[i][j] != '*')
				{
					count = 0;
					int op[] = { 0, 1, -1 };
					for (int t1 = 0; t1 < 3; t1++)
					{
						for (int t2 = 0; t2 < 3; t2++)
						{
							if (i + op[t1] >= 0 && j + op[t2] >= 0 &&
								i + op[t1] < n && j + op[t2] < m &&
								arr[i + op[t1]][j + op[t2]] == '*')
								count++;
						}
					}

					arr[i][j] = trans(count);
				}
			}
		}

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

		for (i = 0; i < n; i++)
		{
			for (j = 0; j < m; j++)
			{
				cout << arr[i][j];
			}
			cout << endl;
		}
		cout << endl;
	}

	return EXIT_SUCCESS;
}