还有螺柱说的求孤岛的，我刚开始用了DFS，没加DP，过了80，提示的错误也是数组越界，然后我猜可能是栈溢出，在本地输入了一个1000X1000全是1的数组，结果溢出了，然而时间所剩无几，没有将DFS改为循环，完了写了循环的方式，可以解决1000X1000全是1的情况，但是不知道能不能真的AC，也有人说是第二个的此时用例还是初始化有问题，那可能是吧，下边是我循环的代码
import java.util.ArrayDeque;
import java.util.Scanner;
/*
4
1 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0
*/
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String line = in.nextLine();
int N = Integer.parseInt(line);
String[] split = null;
boolean stand[][] = new boolean[N][N];
boolean flag[][] = new boolean[N][N];
for (int i = 0; i < N; i++) {
line = in.nextLine();
split = line.split(" ");
for (int j = 0; j < N; j++) {
if (split[j].equals("1"))
stand[i][j] = true;
}
}
System.out.println(findTeam(N, stand));
}
public static int findTeam(int N, boolean stand[][]) {
boolean flag[][] = new boolean[N][N];
ArrayDeque<Integer[]> deque = new ArrayDeque<>();
int ret = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (stand[i][j] && !flag[i][j]) {
deque.offer(new Integer[] { i, j });
flag[i][j] = true;
while (!deque.isEmpty()) {
Integer[] site = deque.poll();
int m = site[0];
int n = site[1];
if (m > 0 && stand[m - 1][n] && !flag[m - 1][n]) {
deque.offer(new Integer[] { m - 1, n });
flag[m - 1][n] = true;
}
if (n > 0 && stand[m][n - 1] && !flag[m][n - 1]) {
deque.offer(new Integer[] { m, n - 1 });
flag[m][n - 1] = true;
}
if (n < N - 1 && stand[m][n + 1] && !flag[m][n + 1]) {
deque.offer(new Integer[] { m, n + 1 });
flag[m][n + 1] = true;
}
if (m < N - 1 && stand[m + 1][n] && !flag[m + 1][n]) {
deque.offer(new Integer[] { m + 1, n });
flag[m + 1][n] = true;
}
}
ret++;
}
}
}
return ret;
}
}
