------
## 最终结论
```python
def shortest_common_supersequence(str1, str2):
    def longest_common_subsequence(X, Y):
        m = len(X)
        n = len(Y)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if X[i - 1] == Y[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
        
        # Reconstruct the longest common subsequence
        lcs = []
        while m > 0 and n > 0:
            if X[m - 1] == Y[n - 1]:
                lcs.append(X[m - 1])
                m -= 1
                n -= 1
            elif dp[m - 1][n] > dp[m][n - 1]:
                m -= 1
            else:
                n -= 1
        
        return ''.join(reversed(lcs))

    lcs = longest_common_subsequence(str1, str2)
    
    # Build the shortest common supersequence using LCS
    i, j = 0, 0
    result = []

    for c in lcs:
        while i < len(str1) and str1[i] != c:
            result.append(str1[i])
            i += 1
        while j < len(str2) and str2[j] != c:
            result.append(str2[j])
            j += 1
        result.append(c)
        i += 1
        j += 1

    result.extend(str1[i:])
    result.extend(str2[j:])
    
    return ''.join(result)

# Example usage:
str1 = "abac"
str2 = "cab"
print(shortest_common_supersequence(str1, str2)) # Output: "cabac" or other valid outputs
```