Mineweep（扫雷）
题目描述：
每周一题之2 Mineweep（扫雷）
Minesweeper （扫雷）
PC/UVa IDs: 110102/10189,
Popularity: A,
Success rate: high Level: 1
测试地址：https://vjudge.net/problem/UVA-10189
[问题描述]
Have you ever played Minesweeper? It’s a cute little game which comes within a certain Operating
System which name we can’t really remember. Well, the goal of the game is to find where are all the mines within a M × N field. To help you, the game shows a number in a square which tells you how many mines there are adjacent to that square. For instance, supose the following 4 × 4 field with 2 mines (which are represented by an ‘*’ character):
*...
....
.*..
....
If we would represent the same field placing the hint numbers described above, we would end up
with:
*100
2210
1*10
1110
As you may have already noticed, each square may have at most 8 adjacent squares.
[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 stands for the number of lines and columns of the field respectively.
The next n lines contains exactly m characters and represent the field.
Each safe square is represented by an ‘.’ character (without the quotes) and each mine square
is represented by an ‘*’ character (also without the quotes). The first field line where n = m = 0
represents the end of input and should not be processed.
下面这段话蓝桥官方误导：
[输出] 对于每对整数 i 和 j，按原来的顺序输出 i 和 j，然后输出二者之间的整数中的最大循环节长度。这三个整数应该用单个空格隔开，且在同一行输出。对于读入的每一组数据，在输出中应位于单独的一行。
[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
*/
解析：
1.处理好边界：
2. 遍历每个点的每个方向，操作结课
AC代码:
package december.year18;
import java.util.Scanner;
public class Solution2 {
static int [] x={0,0,1,-1,1,-1,-1,1};
static int [] y={1,-1,0,0,1,-1,1,-1};
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int cindex=1;
while (in.hasNext()) {
int n = in.nextInt();
int m = in.nextInt();
if(m==0&&m==n) {
return;
}
char[][] mat = new char[n][];
for (int i = 0; i < mat.length; i++) {
mat[i]=in.next().toCharArray();
}
int mL=mat[0].length;
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mL; j++) {
if(mat[i][j]!='*') {
for (int k = 0; k < 8; k++) {
int newx=i+x[k];
int newy=j+y[k];
if(newx>=0&&newy>=0&&newx<n&&newy<mL) {
if(mat[newx][newy]=='*') {
if(mat[i][j]=='.') {
mat[i][j]='0';
}
mat[i][j]++;
}
}
}
}
}
}
if(cindex>1)System.out.println();
System.out.println("Field #"+cindex+":");
cindex++;
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[i].length; j++) {
if(mat[i][j]=='.') {
System.out.print('0');
}else {
System.out.print(mat[i][j]);
}
}
System.out.println();
}
}
}
}
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