把圆当成正方形即可

Nice one.Fun. The best bound might be π(L+2).

Indeed not hard, way easier than that disgusting P2

We will prove the bound of $\pi (l+2), l\geq 2$, $2 \pi l,1\leq l \leq 2$, and $2 \pi, l \leq 1$, which is in fact sharp. Lemma 1: If two circles intersect in the plane then the perimeter of the arcs that lie strictly inside their union is larger than the perimeter of a circle that is internally tangent to both circles. Proof: Cut a sector from the center of the internally tangent circle. Now draw the chords bounded by the sector on both circles. It is not hard to see that the sector and chord of the internally tangent circle bound an isosceles triangle and therefor the length of the chord on the larger circle will be larger. Since perimeter is defined as the limit of the sum of chords bounded by many tiny sectors the lemma easily follows. (Image below helps). Now draw the all radical axes of the circles. We will prove by induction that the perimeter of the union of the set of circles is no more than $\pi$ times the space they take up on the line. This is trivial by repeatedly applying lemma $1$ while moving along the radical axes from left to right. You then get the above bound by considering the maximum distance the circles can take up on the line.

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