poj3076 Sudoku DLX
同样是一道dancing links，和poj3074同样是数独问题，只要稍稍改下就能过，我上篇文章里用的是矩阵存储方式，套到这里居然MLE了，只好重新处理，处理过程还是比较麻烦的，终于搞定了，呵呵。这里我给模板化了，能够很容易的处理数独这一类问题
代码
1
/*
DLX解决sudoku问题，转化为MM * NN的精确覆盖问题
*/
2
//
解决LL * LL的Sudoku问题
3
#include
<
stdio.h
>
4
#include
<
math.h
>
5
#include
<
stdlib.h
>
6
#include
<
string
.h
>
7
#define
INF 0x3fffffff
8
#define
LL 16
9
#define
X 2
10
#define
MM (LL * LL * LL)
11
#define
NN (LL * LL * 4)
12
13
const
int
SQ
=
(
int
)(sqrt(
1.0
*
LL));
14
int
N, M;
15
int
cntc[NN
+
X];
//
记录列节点个数
16
int
head;
17
int
num;
18
char
ans[LL
+
X][LL
+
X];
//
存储最后结果
19
20
struct
node{
21
int
l, r, u, d, x, y, n, c;
//
c表示当前节点所在列, ans[x][y] = n
22
void
set
(
int
_l,
int
_r,
int
_u,
int
_d,
int
_x,
int
_y,
int
_n,
int
_c){
23
l
=
_l; r
=
_r; u
=
_u; d
=
_d; x
=
_x, y
=
_y, n
=
_n, c
=
_c;
24
}
25
}f[MM
*
4
+
NN
+
X];
26
27
void
Add(
int
_l,
int
_x,
int
_y,
int
_n,
int
c){
28
f[
++
num].x
=
_x;
29
f[num].y
=
_y;
30
f[num].n
=
_n;
31
f[f[num].l
=
_l].r
=
num;
32
f[f[num].u
=
f[c].u].d
=
num;
33
f[f[num].d
=
f[num].c
=
c].u
=
num;
34
cntc[c]
++
;
35
}
36
void
Addrow(
int
i,
int
j,
int
k){
//
如果ans[i][j] = k
37
int
t
=
(i
-
1
)
*
LL
+
j;
38
Add(num
+
4
, i, j, k, t);
39
Add(num, i, j, k, LL
*
LL
+
(i
-
1
)
*
LL
+
k);
40
Add(num, i, j, k,
2
*
LL
*
LL
+
(j
-
1
)
*
LL
+
k);
41
Add(num, i, j, k,
3
*
LL
*
LL
+
((i
-
1
)
/
SQ
*
SQ
+
(j
+
SQ
-
1
)
/
SQ
-
1
)
*
LL
+
k);
42
}
43
/*
删除第c列
*/
44
void
remove(
int
c){
45
f[f[c].r].l
=
f[c].l;
46
f[f[c].l].r
=
f[c].r;
47
48
int
i, j;
49
for
(i
=
f[c].d; i
!=
c; i
=
f[i].d){
50
for
(j
=
f[i].r; j
!=
i; j
=
f[j].r){
51
f[f[j].d].u
=
f[j].u;
52
f[f[j].u].d
=
f[j].d;
53
cntc[f[j].c]
--
;
54
}
55
}
56
//
printf("remove %d\n", c);
57
}
58
/*
恢复第c列
*/
59
void
resume(
int
c){
60
f[f[c].r].l
=
c;
61
f[f[c].l].r
=
c;
62
63
int
i, j;
64
for
(i
=
f[c].d; i
!=
c; i
=
f[i].d){
65
for
(j
=
f[i].r; j
!=
i; j
=
f[j].r){
66
f[f[j].d].u
=
j;
67
f[f[j].u].d
=
j;
68
cntc[f[j].c]
++
;
69
}
70
}
71
//
printf("resume %d\n", c);
72
}
73
int
dfs(){
74
if
(f[head].r
==
head)
return
1
;
75
int
min
=
INF;
76
int
c, i, j;
77
for
(i
=
f[head].r; i
!=
head; i
=
f[i].r){
78
if
(cntc[i]
<
min){
79
c
=
i;
80
min
=
cntc[i];
81
}
82
}
83
remove(c);
84
for
(i
=
f[c].d; i
!=
c; i
=
f[i].d){
85
ans[f[i].x][f[i].y]
=
f[i].n;
//
每选中一行相当于填一小格
86
87
for
(j
=
f[i].r; j
!=
i; j
=
f[j].r){
88
remove(f[j].c);
89
}
90
if
(dfs())
return
1
;
91
/*
这个顺序很重要，删除和恢复的方向必须相反
92
开始相同，都是向右的，结果TLE了
*/
93
for
(j
=
f[i].l; j
!=
i; j
=
f[j].l){
94
resume(f[j].c);
95
}
96
}
97
resume(c);
98
return
0
;
99
}
100
101
int
main()
102
{
103
int
i, j, k;
104
char
s[LL
+
X][LL
+
X];
105
int
ca
=
0
;
106
while
(scanf(
"
%s
"
, s[
0
])
!=
EOF){
107
if
(ca) puts(
""
);
108
ca
++
;
109
for
(i
=
1
; i
<
LL; i
++
){
110
scanf(
"
%s
"
, s[i]);
111
}
112
M
=
MM;
113
N
=
NN;
114
num
=
N;
115
for
(i
=
1
; i
<
N; i
++
){
116
f[i].
set
(i
-
1
, i
+
1
, i, i,
0
,
0
,
0
, i);
117
}
118
f[
0
].
set
(N,
1
,
0
,
0
,
0
,
0
,
0
,
0
);
119
f[N].
set
(N
-
1
,
0
, N, N,
0
,
0
,
0
, N);
120
121
memset(cntc,
0
,
sizeof
(cntc));
122
for
(i
=
1
; i
<=
LL; i
++
){
123
for
(j
=
1
; j
<=
LL; j
++
){
124
if
(s[i
-
1
][j
-
1
]
==
'
-
'
){
125
for
(k
=
1
; k
<=
LL; k
++
){
126
Addrow(i, j, k);
127
}
128
}
else
{
129
k
=
s[i
-
1
][j
-
1
]
-
'
A
'
+
1
;
130
Addrow(i, j, k);
131
}
132
}
133
}
134
dfs();
135
for
(i
=
1
; i
<=
LL; i
++
){
136
for
(j
=
1
; j
<=
LL; j
++
) printf(
"
%c
"
, ans[i][j]
+
'
A
'
-
1
);
137
puts(
""
);
138
}
139
}
140
return
0
;
141
}
142