极大团计数模板
极大团的资料已经在前一篇文章中说过了
下了一下官方的数据，有的数据RE了，可能是在本机上系统栈分配的太小的原因吧。
在POJ HOJ都AC了，其中一分用STL写的在 POJ TLE，数据太强了
具体思想如wiki上说的，但是要加上一个强剪枝！！
起初用自己写的bitset，简化了集合求并的操作，但是遍历复杂了，而且没有那个剪枝，所以一直TLE
代码如下：
Bron Kerbosch with STL
1
/*
2
* =====================================================================================
3
*
4
* Filename: clique.cpp
5
*
6
* Description: max clique
7
*
8
* Version: 1.0
9
* Created: 2011年02月23日 15时49分48秒
10
* Revision: none
11
* Compiler: gcc
12
*
13
* Author: ronaflx
14
* Company: hit-acm-group
15
*
16
* =====================================================================================
17
*/
18
#include
<
iostream
>
19
#include
<
sstream
>
20
#include
<
cstring
>
21
#include
<
cstdio
>
22
#include
<
algorithm
>
23
#include
<
vector
>
24
using
namespace
std;
25
class
Bron_Kerbosch
26
{
27
private
:
28
const
static
int
N
=
130
;
29
int
n, maps[N][N], cnt;
30
void
countClique(vector
<
int
>&
p,vector
<
int
>
&
x)
31
{
32
if
(p.size()
==
0
&&
x.size()
==
0
)
33
cnt
++
;
34
for
(vector
<
int
>
::iterator i
=
x.begin(); i
!=
x.end();i
++
)
35
{
36
vector
<
int
>
::iterator j;
37
for
(j
=
p.begin(); j
!=
p.end()
&&
maps[
*
i][
*
j]; j
++
);
38
if
(j
==
p.end())
39
return
;
40
}
//
剪枝！！
41
vector
<
int
>
tmpp, tmpx, copyx;
42
copyx.assign(x.begin(), x.end());
43
for
(vector
<
int
>
::iterator i
=
p.begin(); i
!=
p.end(); i
++
)
44
{
45
tmpp.clear();
46
tmpx.clear();
47
if
(i
!=
p.end()
-
1
)
48
{
49
for
(vector
<
int
>
::iterator j
=
i
+
1
;j
!=
p.end();j
++
)
50
{
51
if
(maps[
*
i][
*
j])
52
tmpp.push_back(
*
j);
53
}
54
}
55
for
(vector
<
int
>
::iterator j
=
copyx.begin();j
!=
copyx.end();j
++
)
56
{
57
if
(maps[
*
i][
*
j])
58
tmpx.push_back(
*
j);
59
}
60
countClique(tmpp, tmpx);
61
if
(cnt
>
1000
)
62
return
;
63
copyx.push_back(
*
i);
64
}
65
x.assign(copyx.begin(), copyx.end());
66
tmpp.clear();
67
tmpx.clear();
68
copyx.clear();
69
}
70
public
:
71
void
initialize(
int
n,
int
m)
72
{
73
memset(maps,
0
,
sizeof
(maps));
74
this
->
n
=
n;
75
for
(
int
i
=
0
;i
<
m;i
++
)
76
{
77
int
a, b;
78
scanf(
"
%d %d
"
,
&
a,
&
b);
79
a
--
, b
--
;
80
maps[a][b]
=
true
;
81
maps[b][a]
=
true
;
82
}
83
}
84
int
countClique()
85
{
86
cnt
=
0
;
87
vector
<
int
>
p, x;
88
for
(
int
i
=
0
;i
<
n;i
++
)
89
p.push_back(i);
90
countClique(p, x);
91
return
cnt;
92
}
93
}one ;
94
int
main()
95
{
96
int
n, m;
97
while
(scanf(
"
%d %d
"
,
&
n,
&
m)
==
2
)
98
{
99
one.initialize(n, m);
100
int
ans
=
one.countClique();
101
if
(ans
>
1000
)
102
printf(
"
Too many maximal sets of friends.\n
"
);
103
else
104
printf(
"
%d\n
"
, ans);
105
}
106
return
0
;
107
}
不用STL的比较高效
Bron Kerbosch
1
#include
<
iostream
>
2
#include
<
sstream
>
3
#include
<
cstring
>
4
#include
<
cstdio
>
5
#include
<
algorithm
>
6
using
namespace
std;
7
class
Bron_Kerbosch
8
{
9
private
:
10
const
static
int
N
=
130
;
11
int
n, maps[N][N], cnt;
12
void
countClique(
int
*
p,
int
ps,
int
*
x,
int
xs)
13
{
14
if
(ps
==
0
)
15
{
16
if
(xs
==
0
)
17
cnt
++
;
18
return
;
19
}
20
for
(
int
i
=
0
;i
<
xs;i
++
)
21
{
22
int
j, v
=
x[i];
23
for
(j
=
0
;j
<
ps
&&
maps[p[j]][v];j
++
);
24
if
(j
==
ps)
25
return
;
26
}
27
int
tmpp[N], tmpps
=
0
, tmpx[N], tmpxs
=
0
;
28
for
(
int
i
=
0
;i
<
ps;i
++
)
29
{
30
int
v
=
p[i];
31
tmpps
=
tmpxs
=
0
;
32
for
(
int
j
=
i
+
1
;j
<
ps;j
++
)
33
{
34
int
u
=
p[j];
35
if
(maps[v][u])
36
tmpp[tmpps
++
]
=
u;
37
}
38
for
(
int
j
=
0
;j
<
xs;j
++
)
39
{
40
int
u
=
x[j];
41
if
(maps[v][u])
42
tmpx[tmpxs
++
]
=
u;
43
}
44
countClique(tmpp, tmpps, tmpx, tmpxs);
45
if
(cnt
>
1000
)
46
return
;
47
x[xs
++
]
=
v;
48
}
49
}
50
public
:
51
void
initialize(
int
n,
int
m)
52
{
53
memset(maps,
0
,
sizeof
(maps));
54
this
->
n
=
n;
55
for
(
int
i
=
0
;i
<
m;i
++
)
56
{
57
int
a, b;
58
scanf(
"
%d %d
"
,
&
a,
&
b);
59
a
--
, b
--
;
60
maps[a][b]
=
true
;
61
maps[b][a]
=
true
;
62
}
63
}
64
int
countClique()
65
{
66
cnt
=
0
;
67
int
p[N], x[N];
68
for
(
int
i
=
0
;i
<
n;i
++
)
69
p[i]
=
i;
70
countClique(p, n, x,
0
);
71
return
cnt;
72
}
73
}one ;
74
int
main()
75
{
76
int
n, m;
77
while
(scanf(
"
%d %d
"
,
&
n,
&
m)
==
2
)
78
{
79
int
a, b;
80
one.initialize(n, m);
81
int
ans
=
one.countClique();
82
if
(ans
>
1000
)
83
printf(
"
Too many maximal sets of friends.\n
"
);
84
else
85
printf(
"
%d\n
"
, ans);
86
}
87
return
0
;
88
}