ijustdo

a = [1,2,3,4,5,6]
x = [(1,3),(2,4),(3,4)]

{1: {3}, 3: {1, 4}, 2: {4}, 4: {2, 3}}
[[6, 5, 4, 1], [3, 2]]

----------------------------
a = [1,2,3,4,5,6,7,8,9]
x = [(1,4),(2,5),(1,5),(5,6),(7,8),(3,9),(2,8),(4,5)]

{1: {4, 5}, 4: {1, 5}, 2: {8, 5}, 5: {1, 2, 4, 6}, 6: {5}, 7: {8}, 8: {2, 7}, 3: {9}, 9: {3}}
[[9, 8, 6, 4], [7, 5, 3], [2, 1]]

看结果没有错
图分割的方法 就。。。。

[code]
a = [1,2,3,4,5,6]
x = [(1,3),(2,4),(3,4)]

a = [1,2,3,4,5,6,7,8,9]
x = [(1,4),(2,5),(1,5),(5,6),(7,8),(3,9),(2,8),(4,5)]

x_info = {}

# 根据互斥信息 找到 每个元素的互斥集合
for i in x:
for j in i:
if j not in x_info:
x_info[j] = set()
x_info[j] = x_info[j] | (set(i) - set([j]))

groups = []
all_is = {}.fromkeys(a)


while a:
ci = a.pop()
rr = []
rr.append(ci)
b = a.copy()
while b:
i = b.pop()
can_in = True
for j in rr:
if i in x_info.get(j, set()):
can_in = False
break
if can_in:
rr.append(i)
a.remove(i)
groups.append(rr)

print(x_info)
print(groups)
[/code]

哈哈 好像我那个不对

@arrow8899 正解 我第一反应也是这么干
可能油更好的方法
但是这个起码可行

a = [1,2,3,4,5,6,7,8,9]
x = [(1,4),(2,5),(1,5),(5,6),(7,8),(3,9),(2,8),(4,5)]

x_info = {}

# 根据互斥信息 找到 每个元素的互斥集合
for i in x:
for j in i:
if j not in x_info:
x_info[j] = set()
x_info[j] = x_info[j] | (set(i) - set([j]))


In [65]: x_info
Out[65]:
{1: {4, 5},
4: {1, 5},
2: {5, 8},
5: {1, 2, 4, 6},
6: {5},
7: {8},
8: {2, 7},
3: {9},
9: {3}}


In [63]: groups = []

In [64]: while a:
...: x = a.pop()
...: rr = []
...: rr.append(x)
...: for i in a:
...: if i not in x_info.get(x, set()):
...: rr.append(i)
...: a.remove(i)
...: groups.append(rr)
...:


In [61]: groups
Out[61]: [[9, 1, 4, 6, 8], [7, 2, 5], [3]]


In [72]: len(groups)
Out[72]: 3

同意 文件描述符未释放 lsof -n|grep 'delete' 应该可以看到
linux 下 不就有个回复删除文件的小技巧么 只要 某个删除文件有进程占用 找到 进程 pid 和 文件 fd
cat /proc/pid/fd/fd_num > /home/xxx
就还原了

都与时俱进弄 web 或者 服务端了

还有小伙 告诉你吧 现在 桌面软件的大多数 都是 他妈 套一个浏览器组件 然后里面 还是做页面的那一套了 哈哈
比纯写桌面快很多 成本也低 写桌面基本就是做页面了

你会发现你无论那个语言都离不开正则
基本现在的编辑器都支持正则吧

没毛病

lsattr chattr 查下这两个干嘛的 你就知道
以前我服务器的重要配置 和有的只读代码 都会 chattr +i