我给出的第一个八皇后的解

  用蛮力法解决，耗了三分钟，一共得到了92个解，不知道有谁有解的个数的确切数字。最近迷python，就用它来实现了，水平有限，请各位高手指点。拿了个人家实现，几乎在1s内就实现了，哎，惭愧阿，我要摆脱蛮力法。
   PS:今天利用闲暇时间想了想，要提高效率，最终就是要给出0-7八个数字的全排列，这样测试空间就会小很多。不过我还没有想出如何给出一个比较高效的全排列。

import time

def display (x):
    i = 0
    while i<8:
        print x[i]
        i+=1

def display_method (method):
    x = [[0 for i in range(8)] for j in range(8)]
    for i in range(8):
        x[method[i][0]][method[i][1]]=1
    display(x)

def stupid_method():
    index=1
    for pos1 in range(8):
        for pos2 in range(8):
            for pos3 in range(8):
                for pos4 in range(8):
                    for pos5 in range(8):
                        for pos6 in range(8):
                            for pos7 in range(8):
                                for pos8 in range(8):
                                    if True ==test(pos1 ,pos2 ,pos3,pos4,pos5,pos6,pos7,pos8):
                                        print "This is "+str(index) + " method"
                                        index+=1

def test(pos1 ,pos2 ,pos3,pos4,pos5,pos6,pos7,pos8) :
    method =([0,pos1],[1,pos2],[2,pos3],[3,pos4],[4,pos5],[5,pos6],[6,pos7],[7,pos8])
    for i in range(8):
        for j in range(i+1,8):
            if attack(method[i],method[j])==True:
                return False
    print "**************************"
    display_method(method)
    return True

def attack(x,y):
    if x[1] == y[1]:
        return True
    if ((x[0]-y[0])**2) == ((x[1]-y[1])**2):
        return True
    return False

x = [[0 for i in range(8)] for j in range(8)]

cur_time1=time.time()
print "Start time "+time.strftime( "%Y-%m-%d %X", time.localtime() )
stupid_method()
cur_time2=time.time()
print "End time "+time.strftime( "%Y-%m-%d %X", time.localtime() )
print "Waste time "+str(cur_time2-cur_time1)+" seconds"