Since $ASHY$ is a square and $AY=5$, we have $AS=SY=5\sqrt{2}$. Since $ABDC$ is a square and $AB=1$, we have $AC=\sqrt{2}$, so $CY=5\sqrt{2}-\sqrt{2}=4\sqrt{2}$. Finally, since $EFHG$ is a square and $EF=1$, we have $EG=GF=1\sqrt{2}$.
[asy] size(6cm); defaultpen(black+1); pair a=(0,5); pair b=(1,5); pair c=(0,4); pair d=(1,4); pair e=(4,1); pair f=(5,1); pair g=(4,0); pair h=(5,0); pair y=(0,0); pair s=(5,5); draw(a--s--h--y--a); draw(c--d--b,gray); draw(g--e--f,gray); draw(d--y--e--s--d); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(y); dot(s); label("$A$",a,NW); label("$B$",b,N); label("$C$",c,W); label("$D$",d,SE); label("$E$",e,NW); label("$F$",f,E); label("$G$",g,S); label("$H$",h,SE); label("$Y$",y,SW); label("$S$",s,NE); label("$1$",(a+b)/2,N); label("$\sqrt{2}$",(c+d)/2,N); label("$1\sqrt{2}$",(e+f)/2,N); label("$5$",(a+s)/2,W); label("$5$",(s+h)/2,E); label("$5\sqrt{2}$",(a+s)/2,NE); label("$5\sqrt{2}$",(s+h)/2,NW); label("$4\sqrt{2}$",(s+y)/2,NW); [/asy]
We can now compute the area of quadrilateral $DYES$ by subtracting the areas of triangles $DYE$ and $YES$ from the area of square $DESY$.
We have $[DYE]=\frac{1}{2}\cdot DY\cdot YE=\frac{1}{2}\cdot(4\sqrt{2})\cdot(5\sqrt{2}-1)=38-2\sqrt{2}$ and $[YES]=\frac{1}{2}\cdot YS\cdot ES=\frac{1}{2}\cdot(5\sqrt{2})\cdot(1\sqrt{2})=\frac{25}{2}$.
The area of square $DESY$ is $(DY+YE)^2=(4\sqrt{2}+5\sqrt{2}-1)^2=100-18\sqrt{2}$. Thus, the area of quadrilateral $DYES$ is \begin{align*}
[DESY]-[DYE]-[YES]&=\left(100-18\sqrt{2}\right)-\left(38-2\sqrt{2}\right)-\left(\frac{25}{2}\right)\
&=61-\frac{49}{2}\sqrt{2}.
\end{align*}Therefore, the area of quadrilateral $DYES$ is $\boxed{61-\frac{49}{2}\sqrt{2}}$.