两道枚举题
http://acm.pku.edu.cn/JudgeOnline/problem?id=2965
我用纯枚举做的，216，用了500多ms
code1
1
//
579MS 2010-05-21 13:57:29
2
#include
<
stdio.h
>
3
#include
<
string
.h
>
4
5
const
int
MK
=
1
<<
16
;
6
//
状态转换
7
int
swt[
20
]
=
{
0
,
8
0xf888
,
0xf444
,
0xf222
,
0xf111
,
9
0x8f88
,
0x4f44
,
0x2f22
,
0x1f11
,
10
0x88f8
,
0x44f4
,
0x22f2
,
0x11f1
,
11
0x888f
,
0x444f
,
0x222f
,
0x111f
12
};
13
14
int
main()
15
{
16
char
ch0[
20
], ch1[
10
];
17
int
I, K, M;
18
int
em
=
0
, min, hd;
19
while
(gets(ch1)) {
20
strcpy(ch0, ch1);
21
I
=
3
;
22
while
(I
--
) {
23
gets(ch1);
24
strcat(ch0, ch1);
25
}
26
int
stu0, stu1;
27
stu0
=
0
;
28
//
生成初始状态
29
for
(I
=
0
; I
<
16
; I
++
) {
30
if
(ch0[I]
==
'
+
'
) stu0
|=
1
<<
(
16
-
I
-
1
);
31
}
32
min
=
0xffffff
;
33
//
枚举
34
for
(em
=
0
; em
<
MK; em
++
) {
35
int
cnt
=
0
;
36
stu1
=
stu0;
37
for
(I
=
0
; I
<
16
; I
++
) {
38
if
( (
1
<<
I)
&
em ) {
39
cnt
++
;
40
stu1
^=
swt[I
+
1
];
41
}
42
}
43
//
stu1 = 0 满足条件
44
if
(
!
stu1) {
45
if
(cnt
<
min) { min
=
cnt; hd
=
em; }
46
}
47
}
48
printf(
"
%d\n
"
, min);
49
for
(I
=
0
; I
<
16
; I
++
) {
50
if
( (
1
<<
I)
&
hd ) {
51
printf(
"
%d %d\n
"
, I
/
4
+
1
, I
%
4
+
1
);
52
}
53
}
54
}
55
return
0
;
56
}
57
再看看No.1的题解
代码
1
可以，下面是我两次AC的代码，前一次算是证明吧
2
Problem:
2965
User: huicpc39
3
Memory: 24K Time: 45MS
4
Language: C Result: Accepted
5
6
Source Code
7
#include
<
stdio.h
>
8
int
main()
9
{
char
s[
5
];
int
i,j,k,st[
4
][
4
];
10
for
(i
=
0
;i
<
4
;
++
i)
for
(j
=
0
;j
<
4
;
++
j)st[i][j]
=
0
;
11
for
(i
=
0
;i
<
4
;
++
i)
12
{scanf(
"
%s
"
,s);
13
for
(j
=
0
;j
<
4
;
++
j)
if
(s[j]
==
'
+
'
)
14
{st[i][j]
++
;
15
for
(k
=
0
;k
<
4
;
++
k)st[k][j]
++
;
16
for
(k
=
0
;k
<
4
;
++
k)st[i][k]
++
;}}
17
for
(i
=
0
;i
<
4
;
++
i)
for
(j
=
0
;j
<
4
;
++
j)st[i][j]
%=
2
;
18
for
(i
=
j
=
0
;i
<
4
;
++
i)
for
(k
=
0
;k
<
4
;
++
k)
if
(st[i][k])j
++
;
19
printf(
"
%d\n
"
,j);
20
for
(i
=
0
;i
<
4
;
++
i)
for
(j
=
0
;j
<
4
;
++
j)
if
(st[i][j])printf(
"
%d %d\n
"
,i
+
1
,j
+
1
);
21
}
22
按照以上思路，直接生成结果。
23
Source Code
24
25
Problem:
2965
User: huicpc39
26
Memory: 16K Time: 0MS
27
Language: C Result: Accepted
28
29
Source Code
30
#include
<
stdio.h
>
31
int
s[
16
]
=
{
0x111f
,
0x222f
,
0x444f
,
0x888f
,
0x11f1
,
0x22f2
,
0x44f4
,
0x88f8
,
32
0x1f11
,
0x2f22
,
0x4f44
,
0x8f88
,
0xf111
,
0xf222
,
0xf444
,
0xf888
};
33
int
main()
34
{
char
c;
int
i,n,j;
35
n
=
i
=
j
=
0
;
36
while
((c
=
getchar())
!=
EOF)
37
if
(c
!=
10
)
if
(c
==
'
+
'
)n
^=
s[i
++
];
else
++
i;
38
for
(i
=
0
;i
<
16
;
++
i)
if
(n
&
(
1
<<
i))
++
j;printf(
"
%d\n
"
,j);
39
for
(i
=
0
;i
<
16
;
++
i)
if
(n
&
(
1
<<
i))printf(
"
%d %d\n
"
,i
/
4
+
1
,i
%
4
+
1
);}
40
http://acm.pku.edu.cn/JudgeOnline/problem?id=1753
枚举第1行的操作，24
然后操作第2-4行，使第i行的操作让i-1行满足全黑或全白。
求出最小的操作数。
code3
1
//
124K 16MS 2010-05-21 10:50:18
2
//
Type:枚举
3
#include
<
stdio.h
>
4
#include
<
string
.h
>
5
6
int
tsf[
20
];
//
保存2^i
7
const
int
MK
=
0xf
;
//
枚举次数
8
//
对每个位置的操作
9
int
swt[
20
]
=
{
0
,
10
0x13
,
0x27
,
0x4E
,
0x8C
,
11
0x131
,
0x272
,
0x4E4
,
0x8C8
,
12
0x1310
,
0x2720
,
0x4E40
,
0x8C80
,
13
0x3100
,
0x7200
,
0xE400
,
0xC800
14
};
15
16
void
init()
17
{
18
int
i;
19
tsf[
1
]
=
1
;
20
for
(i
=
2
; i
<=
16
; i
++
) tsf[i]
=
tsf[i
-
1
]
<<
1
;
21
}
22
23
void
Enum(
int
stu0,
int
&
min)
24
{
25
int
I, K, em, cnt, stu;
26
for
(em
=
0
; em
<=
MK; em
++
) {
27
cnt
=
0
;
28
stu
=
stu0;
29
//
对第1行的操作
30
for
(I
=
1
; I
<=
4
; I
++
) {
31
if
( em
&
tsf[I] ) {
32
stu
^=
swt[I
+
12
];
33
cnt
++
;
34
}
35
}
36
//
对2-4行的操作
37
for
(I
=
16
; I
>
4
; I
--
) {
38
if
( stu
&
tsf[I] ) {
39
stu
^=
swt[I
-
4
];
40
cnt
++
;
41
}
42
}
43
if
(
!
stu) {
44
if
(min
==
-
1
) min
=
cnt;
45
else
if
(cnt
<
min) min
=
cnt;
46
}
47
}
48
}
49
50
int
main()
51
{
52
init();
53
char
ch0[
20
], ch1[
20
];
54
int
I, K;
55
int
stu1, stu2;
//
两个不同的状态
56
int
min1, min2;
57
while
(gets(ch1)) {
58
I
=
3
;
59
strcpy(ch0, ch1);
60
while
(I
--
) {
61
gets(ch1);
62
strcat(ch0, ch1);
63
}
64
stu1
=
0
;
65
stu2
=
0
;
66
//
生成两个初始状态
67
for
(I
=
0
, K
=
16
; I
<
16
; I
++
, K
--
) {
68
if
(ch0[I]
==
'
b
'
) stu1
|=
tsf[K];
69
if
(ch0[I]
==
'
w
'
) stu2
|=
tsf[K];
70
}
71
min1
=
-
1
;
72
Enum(stu1, min1);
73
min2
=
-
1
;
74
Enum(stu2, min2);
75
if
(min1
==
-
1
&&
min2
==
-
1
) puts(
"
Impossible
"
);
76
else
if
(min1
==
-
1
) printf(
"
%d\n
"
, min1);
77
else
if
(min2
==
-
1
) printf(
"
%d\n
"
, min2);
78
else
printf(
"
%d\n
"
, min1
<
min2
?
min1 : min2);
79
}
80
return
0
;
81
}