问题描述(答案148)
　　小蓝有一个 30 行 60 列的数字矩阵，矩阵中的每个数都是 0 或 1 。
110010000011111110101001001001101010111011011011101001111110
010000000001010001101100000010010110001111100010101100011110
001011101000100011111111111010000010010101010111001000010100
101100001101011101101011011001000110111111010000000110110000
010101100100010000111000100111100110001110111101010011001011
010011011010011110111101111001001001010111110001101000100011
101001011000110100001101011000000110110110100100110111101011
101111000000101000111001100010110000100110001001000101011001
001110111010001011110000001111100001010101001110011010101110
001010101000110001011111001010111111100110000011011111101010
011111100011001110100101001011110011000101011000100111001011
011010001101011110011011111010111110010100101000110111010110
001110000111100100101110001011101010001100010111110111011011
111100001000001100010110101100111001001111100100110000001101
001110010000000111011110000011000010101000111000000110101101
100100011101011111001101001010011111110010111101000010000111
110010100110101100001101111101010011000110101100000110001010
110101101100001110000100010001001010100010110100100001000011
100100000100001101010101001101000101101000000101111110001010
101101011010101000111110110000110100000010011111111100110010
101111000100000100011000010001011111001010010001010110001010
001010001110101010000100010011101001010101101101010111100101
001111110000101100010111111100000100101010000001011101100001
101011110010000010010110000100001010011111100011011000110010
011110010100011101100101111101000001011100001011010001110011
000101000101000010010010110111000010101111001101100110011100
100011100110011111000110011001111100001110110111001001000111
111011000110001000110111011001011110010010010110101000011111
011110011110110110011011001011010000100100101010110000010011
010011110011100101010101111010001001001111101111101110011101
　如果从一个标为 1 的位置可以通过上下左右走到另一个标为 1 的位置，则称两个位置连通。与某一个标为 1 的位置连通的所有位置（包括自己）组成一个连通分块。
　　请问矩阵中最大的连通分块有多大？
答案提交
　　这是一道结果填空的题，你只需要算出结果后提交即可。本题的结果为一个整数，在提交答案时只填写这个整数，填写多余的内容将无法得分。

#include<iostream>
#include<algorithm>
using namespace std;
int n, m,res,ans;
char g[65][65];
int dx[]={ 1, 0, -1, 0};
int dy[] = { 0,1,0,-1 };
void dfs(int x, int y)
{

	for (int i = 0; i < 4; i++)
	{
		int a = x + dx[i];
		int b = y + dy[i];
		if (a >= 0 && b >= 0 && a < 30 && b < 60 && g[a][b] == '1')
		{
			res++;
			g[a][b] = '0';
			dfs(a, b);
		}
	}
}
int main()
{
	
	for (int i = 0; i < 30; i++)
		for (int j = 0; j < 60; j++)
			cin >> g[i][j];
	for (int i = 0; i < 30; i++)
		for (int j = 0; j < 60; j++)
		{
			if (g[i][j] == '1') {
				res = 0;
				dfs(i, j);
				ans = max(res, ans);//更新最大值
			}
		}

	cout << ans;
	return 0;
}```