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No. $C_{2k+1}$ is self-centered with $r(G)=d(G)=k$, but obviously does not contain $C_{2k}$ as a subgraph. With regards to the revised question, here is a proof that $G$ contains a cycle of length …

Since there is always an edge between two disjoint cycles, this implies that there are graphs in $\mathcal G_g$ with arbitrarily large clique minors. … Let $C_1, \dots, C_5$ be long cycles and choose a red vertex $r_i$ and a blue vertex $b_i$ on each $C_i$ such that $r_i$ and $b_i$ are far apart on $C_i$. …

The problem is NP-hard, even in the unweighted case (all weights equal to $1$). Indeed, given a graph $G$ and an integer $k$, deciding if $G$ contains an Eulerian subgraph with at least $k$ edges is N …

Entringer and Slater considered this problem in their paper On the Maximum Number of Cycles in a Graph. Let $G$ be a simple connected graph with $m$ edges and $n$ vertices. … It is useful to re-parametrize by letting $d=m-n+1$, and defining $\psi(d)$ to be the maximum number of cycles of a graph with $m-n+1=d$. …

If we are trying to pack vertex disjoint cycles that contain $t$, we might as well include this 'closest' cycle. Now just recurse. …

, $b$ $4$-cycles, and $c$ $6$-cycles. … It follows that the edges of $K_n-M$ can be decomposed into $4$-cycles. By then covering pairs of edges of $M$ with $4$-cycles we get a covering of size $\lceil \binom{n}{2}/4+n/8 \rceil$. …

For a graph $G$, let $\nu(G)$ be the maximum number of edge-disjoint cycles, and let $\tau(G)$ be the minimum size of a set of edges $X$ such that $G-X$ has no cycles. …

One class of graphs for which many NP-hard problems (including finding a Hamiltonian cycle) are easy (linear-time) are graphs of bounded tree-width. Indeed, by Courcelle's theorem any problem which c …

Yes, there are many explicit constructions, although some of them are rather complicated. See this talk of Noga Alon, where he presents a very comprehensive history of the problem. Probably the simp …