[Java] UVa 10189 Minesweeper

題目連結

題意:
踩地雷，測資會給 n*m 的矩陣
*代表地雷，.代表沒事
輸出時也是輸出矩陣
地雷就輸出*
沒事的要輸出自己四面八方的地雷總數

解法:
我的寫法比較土法煉鋼
或許能更快更精簡
寫一個函數，判斷每一格周圍的地雷數
參數是當下的座標，依序判斷八個方位就對了

程式(Java):

	import java.util.*;

	public class Main {
	public static int n, m;

	public static int checkNum(char f[][], int y, int x) {
	if (f[y][x] == '*')
	return 0;

	int sum = 0;
	if (y-1 >= 0 && x-1 >= 0) //leftup
	if (f[y-1][x-1] == '*')
	sum++;
	if (y-1 >= 0) //up
	if (f[y-1][x] == '*')
	sum++;
	if (y-1 >= 0 && x+1 <= m-1) //rightup
	if (f[y-1][x+1] == '*')
	sum++;
	if (x-1 >= 0) //left
	if (f[y][x-1] == '*')
	sum++;
	if (x+1 <= m-1) //right
	if (f[y][x+1] == '*')
	sum++;
	if (y+1 <= n-1 && x-1 >= 0) //leftdown
	if (f[y+1][x-1] == '*')
	sum++;
	if (y+1 <= n-1) //down
	if (f[y+1][x] == '*')
	sum++;
	if (y+1 <= n-1 && x+1 <= m-1) //rightdown
	if (f[y+1][x+1] == '*')
	sum++;
	return sum;
	}

	public static void main(String args[]) {
	Scanner sc = new Scanner(System.in);
	int Cases = 1;
	while (sc.hasNext()) {
	n = Integer.valueOf(sc.next());
	m = Integer.valueOf(sc.next());
	if (n == 0 && m == 0)
	break;

	char f[][] = new char[n][m];
	for (int i = 0; i < n; i++) {
	String s = sc.next();
	for (int j = 0; j < m; j++)
	f[i][j] = s.charAt(j);
	}

	if (Cases != 1)
	System.out.println();
	System.out.println("Field #" + Cases++ + ":");
	for (int i = 0; i < n; i++) {
	for (int j = 0; j < m; j++) {
	if (f[i][j] == '*')
	System.out.print("*");
	else
	System.out.print(checkNum(f, i, j));
	}
	System.out.println();
	}
	}
	}
	}

view raw UVa10189.java hosted with ❤ by GitHub

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