Three, when arranging student dormitories, it is stipulated that two students live in one room. The school's preference order for various combinations of students from grades $A, B, C, D, E$ is
$$
A A, A B, A C, A D, A E, B B, B C, B D,
$$
$B E, C C, C D, C E, D D, D E, E E$.
That is, two students from grade $A$ living together is better than one student from grade $A$ and one student from grade $B$ living together, and so on. Assign values $a, b, c, d, e$ to each grade, such that $a a, a b, a c, \cdots$, follow the same order as (1). For integers $a, b, c, d, e$, find the minimum value of $a+b+c+d+e$ and the values of $a, b, c, d, e$.
Three, Solution: According to the problem, we have
$$
a a>a b>a c>a d>a e>b b>b c>b d>b e>c c>c d
$$
$>c e>d d>d e>e e$, which means
$$
\left\{\begin{array}{l}
a>b>c>d>e, \\
a>\frac{b^{2}}{e}, \\
b>\frac{c^{2}}{e}, \\
c>\frac{d^{2}}{e} .
\end{array}\right.
$$

From (2), (3), and (4), we know that for a given $t$, to make $a+b+c+d+e$ the smallest, $d, c, b, a$ must take the corresponding minimum values, i.e., $d=e+1, c=\left[\frac{d^{2}}{e}\right]+1, b=\left[\frac{c^{2}}{e}\right]+1, a=\left[\frac{b^{2}}{e}\right]+1$ when $a+b+c+d+e$ takes the minimum value.
$$
\begin{array}{l}
c=\left[\frac{d^{2}}{e}\right]+1=\left[\frac{(e+1)^{2}}{e}\right]+1=\left[e+2+\frac{1}{e}\right]+1 \\
=\left\{\begin{array}{l}
e+4=5, e=1 \text { when } \\
e+3 . e>1 \text { when }
\end{array}\right. \\
b=\left[\frac{c^{2}}{e}\right]+1=\left\{\begin{array}{l}
26, \quad e=1, \\
e+7+\left[\frac{9}{e}\right], e>1
\end{array}\right. \\
=\left\{\begin{array}{l}
26, e=1 \\
13,2 \leqslant e \leqslant 5 \\
e+8,6 \leqslant e \leqslant 9 \\
e+7, e \geqslant 10
\end{array}\right. \\
a=\left[\frac{b^{2}}{e}\right]+1=\left\{\begin{array}{l}
677, e=1 \\
{\left[\frac{169}{e}\right]+1,2 \leqslant e \leqslant 5} \\
{\left[\frac{(e+8)^{2}}{e}\right]+1,6 \leqslant e \leqslant 9} \\
{\left[\frac{(e+7)^{2}}{e}\right]+1 . e \geqslant 10}
\end{array}\right. \\
=\left\{\begin{array}{l}
677, e=1 \\
85, e=2 \\
57, e=3 \\
43, e=4 \\
37, e=5 \\
33,6 \leqslant e \leqslant 9 \\
e+9,10 \leqslant e \leqslant 12 \\
\geqslant e+15, e \geqslant 13
\end{array}\right. \\
\end{array}
$$

Therefore, we have $a+b+c+d+e$
$$
=\left\{\begin{array}{l}
711, e=1 \\
108, e=2 \\
83, e=3 \\
72, e=4 \\
69, e=5 \\
69, e=6 \\
4 e+45 \geqslant 73,7 \leqslant e \leqslant 9 \\
\geqslant e+(e+1)+(e+3)+(e+7)+(e+15) \geqslant 76, \\
e \geqslant 10
\end{array}\right.
$$

Therefore, when $a=37, b=13, c=8, d=6, e=5$ or $a=33, b=14, c=9, d=7, e=6$, $a+b+c+d+e$ takes the minimum value of 69.