Three. (25 points) Given that $a, b, c$ are real numbers. The functions are $y_{1} = a x^{2} + b x + c, y_{2} = a x + b (a > 0)$. When $-1 \leqslant x \leqslant 1$, it is given that $-1 \leqslant y_{1} \leqslant 1$ and $y_{2}$ has a maximum value of 2. Try to find the area of the figure formed by connecting in sequence all the lattice points (points with integer coordinates) within and on the boundary of the closed region enclosed by the parabola $y_{1} = a x^{2} + b x + c$ and the line $y_{2} = a x + b$.
Three, from $a>0$, we know that $y_{2}$ increases as $x$ increases.
Therefore, $a+b=2$.
When $x=0,1$, $-1 \leqslant c \leqslant 1, -1 \leqslant a+b+c \leqslant 1$, so $-1 \leqslant c=(a+b+c)-2 \leqslant 1-2 \leqslant-1$.
Thus, $c=-1$. Therefore, when $x=0$, $y_{1}=-1$ is the minimum value of $y_{1}=$ $a x^{2}+b x+c$ in the interval $-1 \leqslant x \leqslant 1$.

Hence, $x=0$ is the axis of symmetry of the function $y_{1}=a x^{2}+b x+c$, i.e., $-\frac{b}{2 a}=0, b=0, a=2$.
Thus, $y_{1}=2 x^{2}-1, y_{2}=2 x$.
Solving the system of equations $\left\{\begin{array}{l}y=2 x^{2}-1, \\ y=2 x\end{array}\right.$, we get
$$
\left\{\begin{array}{l}
x_{1}=\frac{1+\sqrt{3}}{2}, \\
y_{1}=1+\sqrt{3};
\end{array} \left\{\begin{array}{l}
x_{2}=\frac{1-\sqrt{3}}{2}, \\
y_{2}=1-\sqrt{3}.
\end{array}\right.\right.
$$

Therefore, the coordinates of the intersection points of the two graphs are $A\left(\frac{1-\sqrt{3}}{2}, 1-\sqrt{3}\right)$ and $B\left(\frac{1+\sqrt{3}}{2}, 1+\sqrt{3}\right)$. The closed figure formed by the two graphs is shown in Figure 8. Since $\frac{1-\sqrt{3}}{2} \leqslant x \leqslant \frac{1+\sqrt{3}}{2}$ has integer solutions $x=0,1$, the corresponding lattice points are $O(0,0), M(0,-1), N(1,1), P(1,2)$. The quadrilateral $O M N P$ is exactly a parallelogram, with area $S_{\text {quadrilateral }}$ MNP $=$ $1 \times 1=1$.