Rt.
具体而言,对于最小圆覆盖,我们有,若 i,j,k 是原有的 Cres 中三个点,则有 pk∈circ(pk′,pi,pj),这样新的圆能覆盖住原来的所有点,但是对于这题我们没有这样的结论。
例如,当
R77=Circle((231850.00,887100.00),247424.00000000)
R31=Circle((145307.00,741503.00),326852.00000000)
R10=Circle((412747.00,958406.00),197166.00000000)
R2=Circle((672652.00,75788.00),902701.00000000)
thus,our (R2,R31,R77) circle isn't restricted by R77,though we've known it should be restricted when we ends for-loop.that may become serious problems.when we use (R10,R31,R77) to instead (R2,R31,R77),R77 is still useless and the restrict of R2 is thrown away. When R2 is a useful restrict at R[1,77] the process will returned a invalid answer in spite of R2.