Let $$f(x) = \min\{[x-1], [x-2], \ldots, [x-10]\}$$ where $$[t]$$ denotes the greatest integer $$\leq t$$. Then $$\int_0^{10} f(x)dx + \int_0^{10} (f(x))^2 dx + \int_0^{10} |f(x)| dx$$ is equal to _______.

We need to evaluate $$\int_0^{10} f(x)\,dx + \int_0^{10} (f(x))^2\,dx + \int_0^{10} |f(x)|\,dx$$ where $$f(x) = \min\{[x-1], [x-2], \ldots, [x-10]\}$$. Since the floor function $$[t]$$ is non-decreasing in $$t$$ and $$x - 10 \leq x - 9 \leq \ldots \leq x - 1$$, it follows that $$[x-10] \leq [x-9] \leq \ldots \leq [x-1]$$ and hence $$f(x) = [x-10]$$.

On the interval $$[0,10]$$, we have $$x-10\in[-10,0]$$. For $$x\in[k,k+1)$$ with $$k=0,1,\dots,9$$ this means $$x-10\in[k-10,k-9)$$ so that $$[x-10]=k-10$$, and at $$x=10$$ one has $$[x-10]=[0]=0$$.

For the first integral $$I_1=\int_0^{10}f(x)\,dx$$ we sum the constant values of $$f(x)=k-10$$ over each unit interval to obtain $$I_1=\sum_{k=0}^{9}(k-10)\cdot1=\sum_{k=0}^{9}(k-10)=(0+1+\dots+9)-100=45-100=-55\,. $$

Similarly, for $$I_2=\int_0^{10}(f(x))^2\,dx$$ we have $$I_2=\sum_{k=0}^{9}(k-10)^2\cdot1=\sum_{j=1}^{10}j^2=\frac{10\cdot11\cdot21}{6}=385\,. $$

Since $$f(x)\le0$$ on each subinterval, the integral $$I_3=\int_0^{10}|f(x)|\,dx$$ becomes $$I_3=\sum_{k=0}^{9}|k-10|=\sum_{k=0}^{9}(10-k)=10+9+\dots+1=55\,. $$

Combining these results gives $$I_1+I_2+I_3=-55+385+55=385\,,$$ so the required value is $$\boxed{385}$$.