晋太元中,武陵人捕鱼为业。缘溪行,忘路之远近。忽逢桃花林,夹岸数百步,中无杂树,芳草鲜美,落英缤纷。渔人甚异之,复前行,欲穷其林。 林尽水源,便得一山,山有小口,仿佛若有光。便舍船,从口入。初极狭,才通人。复行数十步,豁然开朗。土地平旷,屋舍俨然,有良田、美池、桑竹之属。阡陌交通,鸡犬相闻。其中往来种作,男女衣着,悉如外人。黄发垂髫,并怡然自乐。 见渔人,乃大惊,问所从来。具答之。便要还家,设酒杀鸡作食。村中闻有此人,咸来问讯。自云先世避秦时乱,率妻子邑人来此绝境,不复出焉,遂与外人间隔。问今是何世,乃不知有汉,无论魏晋。此人一一为具言所闻,皆叹惋。余人各复延至其家,皆出酒食。停数日,辞去。此中人语云:“不足为外人道也。”(间隔 一作:隔绝) 既出,得其船,便扶向路,处处志之。及郡下,诣太守,说如此。太守即遣人随其往,寻向所志,遂迷,不复得路。 南阳刘子骥,高尚士也,闻之,欣然规往。未果,寻病终。后遂无问津者。
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#! /usr/bin/python2.7
"""N queens problem.
The (well-known) problem is due to Niklaus Wirth.
This solution is inspired by Dijkstra (Structured Programming). It is
a classic recursive backtracking approach.
"""
N = 8 # Default; command line overrides
class Queens:
def __init__(self, n=N):
self.n = n
self.reset()
def reset(self):
n = self.n
self.y = [None] * n # Where is the queen in column x
self.row = [0] * n # Is row[y] safe?
self.up = [0] * (2*n-1) # Is upward diagonal[x-y] safe?
self.down = [0] * (2*n-1) # Is downward diagonal[x+y] safe?
self.nfound = 0 # Instrumentation
def solve(self, x=0): # Recursive solver
for y in range(self.n):
if self.safe(x, y):
self.place(x, y)
if x+1 == self.n:
self.display()
else:
self.solve(x+1)
self.remove(x, y)
def safe(self, x, y):
return not self.row[y] and not self.up[x-y] and not self.down[x+y]
def place(self, x, y):
self.y[x] = y
self.row[y] = 1
self.up[x-y] = 1
self.down[x+y] = 1
def remove(self, x, y):
self.y[x] = None
self.row[y] = 0
self.up[x-y] = 0
self.down[x+y] = 0
silent = 0 # If true, count solutions only
def display(self):
self.nfound = self.nfound + 1
if self.silent:
return
print '+-' + '--'*self.n + '+'
for y in range(self.n-1, -1, -1):
print '|',
for x in range(self.n):
if self.y[x] == y:
print "Q",
else:
print ".",
print '|'
print '+-' + '--'*self.n + '+'
def main():
import sys
silent = 0
n = N
if sys.argv[1:2] == ['-n']:
silent = 1
del sys.argv[1]
if sys.argv[1:]:
n = int(sys.argv[1])
q = Queens(n)
q.silent = silent
q.solve()
print "Found", q.nfound, "solutions."
if __name__ == "__main__":
main()