[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.
[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
*/
我的输入输出和题目要求略有不同只是为了练习其中的算法，没有达到完全一致。
package 蓝桥杯;
import java.util.Scanner;
public class 扫雷 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int[][] num = new int[100][2];
int k = 0;
while(true) {
System.out.print("请输入两个以空格分开的正整数作为棋盘大小：");
int n = 0,m = 0;
try {
n = in.nextInt();
m = in.nextInt();
}
catch(Exception e) {
System.out.println("整数信息输入错误，请重新输入");
in.nextLine();
continue;
}
if(n == 0 && m == 0) {
System.out.println("扫雷完毕！");
break;
}
if(n < 2 || m <2) {
System.out.println("最小默认棋盘为2x2!!,请重新输入!");
continue;
}
char[][] flag = new char[n][m];
char[][] res = new char[n][m];
//初始化记录数组
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
res[i][j] = '0';
}
}
System.out.println("请输入棋盘(中间不要添加任何分隔符,每行结束用enter换行):");
//完成对地图的存储打印
for(int i = 0;i < n;i++) {
String str = in.next();
if(str.length() != m) {
System.out.println("棋盘信息输入错误!请重新输入此行");
i--;
continue;
}
for(int j = 0;j < m;j++) {
flag[i][j] = str.charAt(j);
}
}
//开始扫雷
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
if(flag[i][j] == '*') {
res[i][j] = '*';
find(i,j,n-1,m-1,res);
}
}
}
//打印记录数组
for(int i = 0;i < n;i++) {
for(int j = 0;j < m;j++) {
System.out.print(res[i][j]+" ");
}
System.out.println();
}
System.out.println("-----------------------------------------------------------------");
}
}
public static void find(int i,int j,int n,int m,char[][] res) {
//对四个顶点做特殊处理
if(i == 0 && j == 0) {
add(i,j+1,res);
add(i+1,j+1,res);
add(i+1,j,res);
}
else if(i == 0 && j == m) {
add(i,j-1,res);
add(i+1,j-1,res);
add(i+1,j,res);
}
else if(i == n && j == 0) {
add(i-1,j,res);
add(i-1,j+1,res);
add(i,j+1,res);
}
else if(i == n && j == m) {
add(i,j-1,res);
add(i-1,j-1,res);
add(i-1,j,res);
}
//对于四条临界边做特殊处理
else if(i == 0) {
add(i,j-1,res);
add(i+1,j-1,res);
add(i+1,j,res);
add(i+1,j+1,res);
add(i,j+1,res);
}
else if(j == 0) {
add(i-1,j,res);
add(i-1,j+1,res);
add(i,j+1,res);
add(i+1,j+1,res);
add(i+1,j,res);
}
else if(i == n) {
add(i,j-1,res);
add(i-1,j-1,res);
add(i-1,j,res);
add(i-1,j+1,res);
add(i,j+1,res);
}
else if(j == m) {
add(i-1,j,res);
add(i-1,j-1,res);
add(i,j-1,res);
add(i+1,j-1,res);
add(i+1,j,res);
}
//对于一般情况做出处理
else {
add(i-1,j-1,res);
add(i-1,j,res);
add(i-1,j+1,res);
add(i,j+1,res);
add(i+1,j+1,res);
add(i+1,j,res);
add(i+1,j-1,res);
add(i,j-1,res);
}
}
public static void add(int i,int j,char[][] res) {
if(res[i][j] != '*')
res[i][j] = (char)(res[i][j]+1);
else
return;
}
}