USACO 1.5 Checker Challenge
题目描述：n皇后问题，规模6－13，按顺序输出前三组解，再输出总解数。
前一段学了点DLX的皮毛，就用它做了做，谁知道前三组解得输出顺序老出错。于是就放弃了。
看了hint，就暴搜了一回，开了4个数组，记录行、列、'/'，'\'，的放皇后的情况，根据对称性，搜一半就行了。
my code
/*
ID: superbi1
LANG: C
TASK: checker
*/
#include
<
stdio.h
>
#include
<
string
.h
>
#include
<
stdlib.h
>
#include
<
time.h
>
#define
NL 15
int
fr[NL][
2
], fc[NL][
2
], d1[NL
*
2
][
2
], d2[NL
*
2
][
2
];
int
col[NL];
int
n, cnt, re, flg;
FILE
*
fout;
void
DFS(
int
r)
{
int
I;
if
(r
==
n
+
1
) {
if
(n
>
6
&&
col[
1
]
>
(n
+
1
)
/
2
) { flg
=
0
;
return
; }
if
(cnt
<
3
) {
for
(I
=
1
; I
<=
n; I
++
) {
if
(I
>
1
) fprintf(fout,
"
"
);
fprintf(fout,
"
%d
"
, col[I]);
}
fprintf(fout,
"
\n
"
);
}
if
(n
&
1
&&
col[
1
]
==
(n
+
1
)
/
2
) re
++
;
cnt
++
;
return
;
}
for
(I
=
1
; I
<=
n; I
++
) {
if
(
!
flg)
return
;
if
(
!
fr[r][
1
]
&&
!
fc[I][
1
]
&&
!
d1[r
-
I
+
n][
1
]
&&
!
d2[r
+
I][
1
]) {
fr[r][
1
]
=
1
;
fc[I][
1
]
=
1
;
d1[r
-
I
+
n][
1
]
=
1
;
d2[r
+
I][
1
]
=
1
;
fr[r][
0
]
++
;
fc[I][
0
]
++
;
d1[r
-
I
+
n][
0
]
++
;
d2[r
+
I][
0
]
++
;
col[r]
=
I;
DFS(r
+
1
);
fr[r][
0
]
--
;
fc[I][
0
]
--
;
d1[r
-
I
+
n][
0
]
--
;
d2[r
+
I][
0
]
--
;
if
(fr[r][
0
]
==
0
) fr[r][
1
]
=
0
;
if
(fc[I][
0
]
==
0
) fc[I][
1
]
=
0
;
if
(d1[r
-
I
+
n][
0
]
==
0
) d1[r
-
I
+
n][
1
]
=
0
;
if
(d2[r
+
I][
0
]
==
0
) d2[r
+
I][
1
]
=
0
;
}
}
}
int
main()
{
int
time
=
clock();
FILE
*
fin
=
fopen(
"
checker.in
"
,
"
r
"
);
fout
=
fopen(
"
checker.out
"
,
"
w
"
);
fscanf(fin,
"
%d
"
,
&
n);
memset(fr,
0
,
sizeof
(fr));
memset(fc,
0
,
sizeof
(fc));
memset(d1,
0
,
sizeof
(d1));
memset(d2,
0
,
sizeof
(d2));
cnt
=
0
;
re
=
0
;
flg
=
1
;
DFS(
1
);
if
(n
==
6
) cnt
=
2
;
fprintf(fout,
"
%d\n
"
, (cnt
-
re)
*
2
+
re);
printf(
"
%dms\n
"
, clock()
-
time);
return
0
;
}
看了解题报告，发现自己写的好烂，首先，不需要记录行（why?），其次，直接记录个数，标记是多余的。
standard
1
/*
2
TASK: checker
3
LANG: C
4
*/
5
#include
<
stdio.h
>
6
#include
<
stdlib.h
>
7
#include
<
string
.h
>
8
#include
<
assert.h
>
9
10
#define
MAXN 16
11
12
int
n;
13
int
nsol, nprinted;
14
char
row[MAXN];
15
FILE
*
fout;
16
17
void
18
solution() {
19
int
i;
20
for
(i
=
0
; i
<
n; i
++
) {
21
if
(i
!=
0
) fprintf(fout,
"
"
);
22
fprintf(fout,
"
%d
"
, row[i]
+
1
);
23
}
24
fprintf(fout,
"
\n
"
);
25
}
26
27
/*
Keep track of whether there is a checker on each column, and diagonal.
*/
28
char
col[MAXN];
/*
(i, j) -> j
*/
29
char
updiag[
2
*
MAXN];
/*
(i, j) -> i+j
*/
30
char
downdiag[
2
*
MAXN];
/*
(i, j) -> i-j + N
*/
31
32
/*
33
* Calculate number of ways to place checkers
34
* on each row of the board starting at row i and going to row n.
35
*/
36
void
37
nway(i, lim) {
38
int
j;
39
40
if
(i
==
n) {
41
nsol
++
;
42
if
(n
>
6
&&
row[
0
]
<
n
/
2
) nsol
++
;
43
if
(nprinted
++
<
3
) solution();
44
return
;
45
}
46
47
for
(j
=
0
; j
<
lim; j
++
){
48
if
(
!
col[j]
&&
!
updiag[i
+
j]
&&
!
downdiag[i
-
j
+
MAXN]){
49
row[i]
=
j;
50
51
col[j]
++
;
52
updiag[i
+
j]
++
;
53
downdiag[i
-
j
+
MAXN]
++
;
54
55
nway(i
+
1
,n);
56
57
col[j]
--
;
58
updiag[i
+
j]
--
;
59
downdiag[i
-
j
+
MAXN]
--
;
60
}
61
}
62
}
63
64
main(
void
) {
65
FILE
*
fin
=
fopen(
"
checker.in
"
,
"
r
"
);
66
fout
=
fopen(
"
checker.out
"
,
"
w
"
);
67
fscanf(fin,
"
%d
"
,
&
n);
68
nway(
0
, n
>
6
?
(n
+
1
)
/
2
:n);
69
fprintf(fout,
"
%d\n
"
, nsol);
70
exit (
0
);
71
}
72
73
这个更牛，用的位标记+栈模拟的递归，有空再研究吧~
牛叉
1
#include
<
stdio.h
>
2
#include
<
stdlib.h
>
3
#include
<
fstream.h
>
4
#define
MAX_BOARDSIZE 16
5
typedef unsigned
long
SOLUTIONTYPE;
6
#define
MIN_BOARDSIZE 6
7
SOLUTIONTYPE g_numsolutions
=
0
;
8
9
void
Nqueen(
int
board_size) {
10
int
aQueenBitRes[MAX_BOARDSIZE];
/*
results
*/
11
int
aQueenBitCol[MAX_BOARDSIZE];
/*
marks used columns
*/
12
int
aQueenBitPosDiag[MAX_BOARDSIZE];
/*
marks used "positive diagonals"
*/
13
int
aQueenBitNegDiag[MAX_BOARDSIZE];
/*
marks used "negative diagonals"
*/
14
int
aStack[MAX_BOARDSIZE
+
2
];
/*
a stack instead of recursion
*/
15
int
*
pnStack;
16
17
int
numrows
=
0
;
/*
numrows redundant - could use stack
*/
18
unsigned
int
lsb;
/*
least significant bit
*/
19
unsigned
int
bitfield;
/*
set bits denote possible queen positions
*/
20
int
i;
21
int
odd
=
board_size
&
1
;
/*
1 if board_size odd
*/
22
int
board_m1
=
board_size
-
1
;
/*
board size - 1
*/
23
int
mask
=
(
1
<<
board_size)
-
1
;
/*
N bit mask of all 1's
*/
24
25
aStack[
0
]
=
-
1
;
/*
sentinel signifies end of stack
*/
26
for
(i
=
0
; i
<
(
1
+
odd);
++
i) {
27
bitfield
=
0
;
28
if
(
0
==
i) {
29
int
half
=
board_size
>>
1
;
/*
divide by two
*/
30
bitfield
=
(
1
<<
half)
-
1
;
31
pnStack
=
aStack
+
1
;
/*
stack pointer
*/
32
aQueenBitRes[
0
]
=
0
;
33
aQueenBitCol[
0
]
=
aQueenBitPosDiag[
0
]
=
aQueenBitNegDiag[
0
]
=
0
;
34
}
else
{
35
bitfield
=
1
<<
(board_size
>>
1
);
36
numrows
=
1
;
/*
prob. already 0
*/
37
38
aQueenBitRes[
0
]
=
bitfield;
39
aQueenBitCol[
0
]
=
aQueenBitPosDiag[
0
]
=
aQueenBitNegDiag[
0
]
=
0
;
40
aQueenBitCol[
1
]
=
bitfield;
41
42
aQueenBitNegDiag[
1
]
=
(bitfield
>>
1
);
43
aQueenBitPosDiag[
1
]
=
(bitfield
<<
1
);
44
pnStack
=
aStack
+
1
;
/*
stack pointer
*/
45
*
pnStack
++
=
0
;
/*
row done -- only 1 element & we've done it
*/
46
bitfield
=
(bitfield
-
1
)
>>
1
;
47
/*
bitfield -1 is all 1's to the left of the single 1
*/
48
}
49
for
(;;) {
50
lsb
=
-
((signed)bitfield)
&
bitfield;
51
/*
this assumes a 2's complement architecture
*/
52
if
(
0
==
bitfield) {
53
bitfield
=
*--
pnStack;
/*
get prev. bitfield from stack
*/
54
if
(pnStack
==
aStack)
/*
if sentinel hit....
*/
55
break
;
56
--
numrows;
57
continue
;
58
}
59
bitfield
&=
~
lsb;
/*
bit off -> don't try it again
*/
60
aQueenBitRes[numrows]
=
lsb;
/*
save the result
*/
61
if
(numrows
<
board_m1) {
/*
more rows to process?
*/
62
int
n
=
numrows
++
;
63
aQueenBitCol[numrows]
=
aQueenBitCol[n]
|
lsb;
64
aQueenBitNegDiag[numrows]
=
(aQueenBitNegDiag[n]
|
lsb)
>>
1
;
65
aQueenBitPosDiag[numrows]
=
(aQueenBitPosDiag[n]
|
lsb)
<<
1
;
66
*
pnStack
++
=
bitfield;
67
bitfield
=
mask
&
~
(aQueenBitCol[numrows]
|
68
aQueenBitNegDiag[numrows]
|
aQueenBitPosDiag[numrows]);
69
continue
;
70
}
else
{
71
++
g_numsolutions;
72
bitfield
=
*--
pnStack;
73
--
numrows;
74
continue
;
75
}
76
}
77
}
78
g_numsolutions
*=
2
;
79
}
80
81
int
main(
int
argc,
char
**
argv) {
82
ifstream f(
"
checker.in
"
);
83
ofstream g(
"
checker.out
"
);
84
long
boardsize,s[
20
],ok,k,i,sol
=
0
;
85
f
>>
boardsize;
86
Nqueen (boardsize);
87
k
=
1
;
88
s[k]
=
0
;
89
while
(k
>
0
) {
90
ok
=
0
;
91
while
(s[k]
<
boardsize
&&
!
ok) {
92
ok
=
1
;
93
s[k]
++
;
94
for
(i
=
1
;i
<
k;i
++
)
95
if
(s[i]
==
s[k]
||
abs(s[k]
-
s[i])
==
abs(k
-
i))
96
ok
=
0
;
97
}
98
if
(sol
!=
3
)
99
if
(
!
ok)
100
k
--
;
101
else
102
if
(k
==
boardsize) {
103
for
(i
=
1
;i
<
boardsize;i
++
) {
104
for
(
int
j
=
1
;j
<=
boardsize;j
++
)
105
if
(s[i]
==
j) g
<<
j
<<
"
"
;
106
}
107
for
(i
=
1
;i
<=
boardsize;i
++
)
108
if
(s[boardsize]
==
i) g
<<
i;
109
g
<<
"
\n
"
;
110
sol
++
;
111
}
else
{
112
k
++
;
113
s[k]
=
0
;
114
}
else
break
;
115
}
116
g
<<
g_numsolutions
<<
"
\n
"
;
117
f.close();
118
g.close();
119
return
0
;
120
}
121
122