MM26 Python Solution (check time every times)

Permute.py

import math, sys, time

class Permute:
    limit = 29.8
    def findOrder(self, w):
        time0 = time.time() + self.limit
        n = int(math.sqrt(len(w)))
        ret = range(n)
        sc = 0.0
        for i in xrange(n - 1):
            rn = ret[i] * n
            for r in ret[i + 1:]:
                sc += w[rn + r]
        lp, a, b = 0, 0, 1
        while 1:
            time1 = time.time()
            if time1 > time0:
                break
            lp += 1
            ra, rb = ret[a], ret[b]
            ran, rbn = ra * n, rb * n
            dsc = w[rbn + ra] - w[ran + rb]
            for r in ret[a + 1:b]:
                rn = r * n
                dsc += w[rbn + r] + w[rn + ra] - w[ran + r] - w[rn + rb]
            if dsc > 0.0:
                ret[a], ret[b] = rb, ra
                sc += dsc
            b += 1
            if b >= n:
                a += 1
                if a >= n - 1:
                    a = 0
                b = a + 1
        dv = n * math.sqrt(n * math.log(n) - n) / 100.0
        sc /= dv
        sys.stderr.write("score={}\n".format(sc))
        sys.stderr.write("loop={}\n".format(lp))
        return tuple(ret)