------
## 最终结论
```python
def max_unique_substrings(s):
    def backtrack(start, seen):
        if start == len(s):
            return 0
        
        max_count = 0
        for end in range(start + 1, len(s) + 1):  
            substring = s[start:end]
            if substring not in seen:  
                seen.add(substring)
                max_count = max(max_count, 1 + backtrack(end, seen))
                seen.remove(substring)
        
        return max_count

    return backtrack(0, set())
```