依次处理每个颜色序号为.的格子，从颜色1往后枚举到一个最小的不会重复的颜色

#include<bits/stdc++.h>
using namespace std;
const int N = 710;
char ch[N][N];
int main()
{
	int h, w; cin >> h >> w;
	for (int i = 1; i <= h; i++)
		for (int j = 1; j <= w; j++)
			cin >> ch[i][j];
	for (int i = 1; i <= h; i++)
		for (int j = 1; j <= w; j++)
		{
			if (ch[i][j] != '.') continue;
			int a = ch[i - 1][j] - '0', b = ch[i][j - 1] - '0',
				c = ch[i + 1][j] - '0', d = ch[i][j + 1] - '0';
			int id = 1;
			while (id == a or id == b or id == c or id == d) id++;
			ch[i][j] = id + '0';
		}
	for (int i = 1; i <= h; i++)
	{
		for (int j = 1; j <= w; j++) cout << ch[i][j];
		cout << '\n';
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
char a[910][910];
int main()
{
	int n,m;cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cin>>a[i][j];
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(a[i][j]=='.')
			{
				if(a[i+1][j]!='1'&&a[i-1][j]!='1'&&a[i][j+1]!='1'&&a[i][j-1]!='1'){a[i][j]='1';continue;}
				if(a[i+1][j]!='2'&&a[i-1][j]!='2'&&a[i][j+1]!='2'&&a[i][j-1]!='2'){a[i][j]='2';continue;}
				if(a[i+1][j]!='3'&&a[i-1][j]!='3'&&a[i][j+1]!='3'&&a[i][j-1]!='3'){a[i][j]='3';continue;}
				if(a[i+1][j]!='4'&&a[i-1][j]!='4'&&a[i][j+1]!='4'&&a[i][j-1]!='4'){a[i][j]='4';continue;}
				if(a[i+1][j]!='5'&&a[i-1][j]!='5'&&a[i][j+1]!='5'&&a[i][j-1]!='5'){a[i][j]='5';continue;}//判断四周的方块
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cout<<a[i][j];
		}
		puts("");
	}
}

#include <bits/stdc++.h>
using namespace std;

const int N = 705;
constexpr int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

int n, m;
char c[N][N];

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) cin >> c[i][j];
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) if (c[i][j] == '.') {
		set <int> s;
		for (int k = 0; k < 4; ++k) {
			int di = i + dir[k][0], dj = j + dir[k][1];
			if (di && di <= n && dj && dj <= m && c[di][dj] != '.') s.insert(c[di][dj]);
		}
		for (int k = '1', flag = 0; k <= '5' && !flag; ++k) if (s.find(k) == s.end()) flag = 1, c[i][j] = k;
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) cout << c[i][j];
		if (i < n) cout << '\n';
	}
	return 0;
}