Examples of manifolds with effective circle actions?
As already pointed out, whenever you have a compact group $G$ acting on a manifold $M$, this action can be made isometric by constructing a metric on $M$ via a standard averaging process. In particular, if you have a (effective) circle action on $M$, you can find a metric $g$ on $M$ such that this action is isometric. In other words, the isometry group of $(M,g)$ contains a circle. This should provide you with plenty of examples: intuitively, a manifold that carries a circle action is one that has (at least) a "circle" worth of symmetries.
For example, take $M$ to be any rotationally symmetric surface in $\mathbb R^3$, e.g., a round sphere $S^2$ or a torus $S^1\times S^1$. Then $M$, by definition, has an isometric (effective) circle action. Moreover, these are the only (orientable) manifolds of dimension $2$ with an effective circle action. This follows from the fact that if we have a torus action on $M$, then the Euler characteristic of $M$ is $\chi(M)=\chi(M^T)$, where $M^T$ is the fixed point set of the torus. Since $M^T$ is a totally geodesic submanifold of $M$, it can be a circle or a finite collection of points. Thus, $\chi(M)\geq 0$, which means $M$ is homeomorphic to $S^2$, $\mathbb RP^2$, $S^1\times S^1$ or the Klein bottle $K^2$, according to orientability and $\chi(M)=2,1,0$, respectively. This is the classification you ask for in dimension equal to $2$.
Classifying compact connected manifolds of any dimension with a circle action and nothing else seems rather too general. More interesting problems are, for example, classifying manifolds with an isometric circle action that also have, e.g., some curvature condition. For example, if $\dim M=4$ and $M$ has a metric of positive curvature that has an isometric circle action, then Hsiang-Kleiner proved that $M$ must be homeomorphic to $S^4$ or $\mathbb C P^2$.
It might be worth pointing out a few non-examples as well; that is, manifolds that do not admit any effective $S^1$-actions (besides surfaces of genus $\geq 2$). It was proved by Borel that a closed aspherical manifold with centerless fundamental group does not admit nontrivial actions by a circle (hence any compact connected Lie group). Moreover, Atiyah and Hirzebruch showed that a closed spin manifold with non-vanishing $\hat A$-genus does not admit non-trivial circle actions.
In the answer of Will, one can exchange manifold $N$ to an orbifold with singularities modeled on $\mathbb R^{n-1}/\mathbb Z_k$ and $\mathbb R^{n}/\mathbb S^1$. Then his answer works for effective action. You can get a lot of examples this way, but I do not think you can call it a "classification".
On the other hand assume you have a manifold $M$ and want to know if it admits an effective $\mathbb S^1$-action. In this case Will's answer is useless and as far as I know no good answer is known.
If there is an action on $M$ then
- There are some restrictions on $\pi_1(M)$; for example compact hyperbolic manifolds do not admit smooth $\mathbb S^1$-actions;
- If $M$ is simply connected and spin then $\hat A(M)$ has to vanish.
I do not know any other restriction.
Assuming you mean: An action of the group $\mathbb R/\mathbb Z$ without stabilizer:
If you look at the quotient of the manifold $M$ by the action, you get a manifold $N$ with a circle bundle $M$. Circle bundles are classified by their Chern class, which is an element of $H^2(N,\mathbb Z)$ coming from the exact sequence $H^1(\mathbb Z) \to H^1(\mathbb R) \to H^1(\mathbb R/\mathbb Z) \to H^2(\mathbb Z)$.
To find a volume preserved by the circle action, take any volume and average it with its pullbacks along the circle action.
So the classification is any smooth compact connected manifold $N$ plus an element of $H^2(N,\mathbb Z)$.