hdu 1041 Solitaire
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1068 Accepted Submission(s): 344
本题对本人来说是个挑战，花费了一天的时间，总算搞定，本题的思想就是 双向BFS+MAP+位压缩存储。声明两个队列，分别表示正向搜索，和反向搜索，再声明两个MAP分别存储两个队列搜索过的状态，由于本题有8行8列，所以用位压缩来存储状态，用unsigned __int64整数表示(__int64位恐怕不行）。两个MAP有两个作用：
第一：当当前队列在本MAP中找到相同状态时表示已经搜过，不用再搜；
第二：当当前队列在对方MAP中找到相同状态时表示找到，可以到达；
代码：
1
#include
<
stdio.h
>
2
#include
<
math.h
>
3
#include
<
map
>
4
#include
<
queue
>
5
using
namespace
std;
6
typedef
struct
node
7
{
8
int
x,y;
9
}NODE;
10
typedef
struct
m
11
{
12
NODE s[
4
];
13
int
time;
14
}MAP;
15
int
main()
16
{
17
map
<
unsigned __int64,
int
>
m1;
18
map
<
unsigned __int64,
int
>
m2;
19
queue
<
MAP
>
qu1,qu2;
20
MAP cur1,cur2,next1,next2;
21
unsigned __int64 s;
22
int
sx,sy,ex,ey,i,j,k,mark,map1[
9
][
9
];
23
int
dir[
4
][
2
]
=
{{
1
,
0
},{
-
1
,
0
},{
0
,
1
},{
0
,
-
1
}};
24
while
(scanf(
"
%d%d
"
,
&
sx,
&
sy)
!=
EOF)
25
{
26
m1.clear ();
27
m2.clear ();
28
while
(
!
qu1.empty ())
29
qu1.pop();
30
while
(
!
qu2.empty ())
31
qu2.pop();
32
s
=
0
;mark
=
0
;
33
cur1.s[
0
].x
=
sx;
34
cur1.s[
0
].y
=
sy;
35
s
+=
(unsigned __int64)pow(
2.0
,(
int
)((sx
-
1
)
*
8
+
sy
-
1
));
36
for
(i
=
1
;i
<=
3
;i
++
)
37
{
38
scanf(
"
%d%d
"
,
&
sx,
&
sy);
39
cur1.s[i].x
=
sx;
40
cur1.s[i].y
=
sy;
41
s
+=
(unsigned __int64)pow(
2.0
,(
int
)((sx
-
1
)
*
8
+
sy
-
1
));
42
}
43
cur1.time
=
0
;
44
m1[s]
=
cur1.time;
45
qu1.push(cur1);
46
s
=
0
;
47
for
(i
=
0
;i
<
4
;i
++
)
48
{
49
scanf(
"
%d%d
"
,
&
ex,
&
ey);
50
cur2.s[i].x
=
ex;
51
cur2.s[i].y
=
ey;
52
s
+=
(unsigned __int64)pow(
2.0
,((ex
-
1
)
*
8
+
ey
-
1
));
53
}
54
cur2.time
=
0
;
55
map
<
unsigned __int64,
int
>
::iterator it;
56
it
=
m1.find(s);
57
if
(it
!=
m1.end())
58
{
59
printf(
"
YES\n
"
);
60
continue
;
61
}
62
m2[s]
=
cur2.time;
63
qu2.push(cur2);
64
while
(
!
qu1.empty()
||
!
qu2.empty ())
65
{
66
if
(
!
qu2.empty()
&&
(qu1.size ()
>=
qu2.size ()
||
qu1.empty ()))
67
{
68
cur2
=
qu2.front();
69
while
(
!
qu2.empty()
&&
cur2.time
>
4
)
70
{
71
qu2.pop();
72
cur2
=
qu2.front ();
73
}
74
qu2.pop();
75
s
=
0
;
76
for
(i
=
0
;i
<
4
;i
++
)
77
{
78
s
+=
(unsigned __int64)pow(
2.0
,((cur2.s[i].x
-
1
)
*
8
+
cur2.s[i].y
-
1
));
79
}
80
it
=
m1.find(s);
81
if
(it
!=
m1.end() )
82
{
83
if
((
*
it).second
+
cur2.time
<=
8
)
84
{
85
mark
=
1
;
86
break
;
87
}
88
else
89
continue
;
90
}
91
memset(map1,
0
,
sizeof
(map1));
92
for
(i
=
0
;i
<
4
;i
++
)
93
{
94
map1[cur2.s[i].x][cur2.s[i].y]
=
1
;
95
}
96
for
(i
=
0
;i
<
4
;i
++
)
97
{
98
next2
=
cur2;
99
sx
=
cur2.s[i].x;
100
sy
=
cur2.s[i].y;
101
for
(j
=
0
;j
<
4
;j
++
)
102
{
103
ex
=
sx
+
dir[j][
0
];
104
ey
=
sy
+
dir[j][
1
];
105
if
(ex
>=
1
&&
ex
<=
8
&&
ey
>=
1
&&
ey
<=
8
&&
map1[ex][ey]
==
1
)
106
{
107
ex
=
sx
+
2
*
dir[j][
0
];
108
ey
=
sy
+
2
*
dir[j][
1
];
109
}
110
if
(ex
>=
1
&&
ex
<=
8
&&
ey
>=
1
&&
ey
<=
8
&&
map1[ex][ey]
==
0
)
111
{
112
s
=
0
;
113
next2.s[i].x
=
ex;
114
next2.s[i].y
=
ey;
115
for
(k
=
0
;k
<
4
;k
++
)
116
{
117
s
+=
(unsigned __int64)pow(
2.0
,((next2.s[k].x
-
1
)
*
8
+
next2.s[k].y
-
1
));
118
}
119
it
=
m2.find(s);
120
next2.time
=
cur2.time
+
1
;
121
if
(next2.time
>
4
)
122
continue
;
123
if
(it
!=
m2.end()
&&
(
*
it).second
<=
next2.time)
124
continue
;
125
m2[s]
=
next2.time;
126
qu2.push(next2);
127
}
128
}
129
}
130
}
131
else
132
{
133
cur1
=
qu1.front ();
134
while
(
!
qu1.empty ()
&&
cur1.time
>
4
)
135
{
136
qu1.pop();
137
cur1
=
qu1.front ();
138
}
139
qu1.pop();
140
s
=
0
;
141
for
(i
=
0
;i
<
4
;i
++
)
142
{
143
s
+=
(unsigned __int64)pow(
2.0
,((cur1.s[i].x
-
1
)
*
8
+
cur1.s[i].y
-
1
));
144
}
145
it
=
m2.find(s);
146
if
(it
!=
m2.end())
147
{
148
if
((
*
it).second
+
cur1.time
<=
8
)
149
{
150
mark
=
1
;
151
break
;
152
}
153
else
154
continue
;
155
}
156
memset(map1,
0
,
sizeof
(map1));
157
for
(i
=
0
;i
<
4
;i
++
)
158
{
159
map1[cur1.s[i].x][cur1.s[i].y]
=
1
;
160
}
161
for
(i
=
0
;i
<
4
;i
++
)
162
{
163
next1
=
cur1;
164
sx
=
cur1.s[i].x;
165
sy
=
cur1.s[i].y;
166
for
(j
=
0
;j
<
4
;j
++
)
167
{
168
ex
=
sx
+
dir[j][
0
];
169
ey
=
sy
+
dir[j][
1
];
170
if
(ex
>=
1
&&
ex
<=
8
&&
ey
>=
1
&&
ey
<=
8
&&
map1[ex][ey]
==
1
)
171
{
172
ex
=
sx
+
2
*
dir[j][
0
];
173
ey
=
sy
+
2
*
dir[j][
1
];
174
}
175
if
(ex
>=
1
&&
ex
<=
8
&&
ey
>=
1
&&
ey
<=
8
&&
map1[ex][ey]
==
0
)
176
{
177
s
=
0
;
178
next1.s[i].x
=
ex;
179
next1.s[i].y
=
ey;
180
for
(k
=
0
;k
<
4
;k
++
)
181
{
182
s
+=
(unsigned __int64)pow(
2.0
,((next1.s[k].x
-
1
)
*
8
+
next1.s[k].y
-
1
));
183
}
184
it
=
m1.find(s);
185
next1.time
=
cur1.time
+
1
;
186
if
(next1.time
>
4
)
187
continue
;
188
if
(it
!=
m1.end()
&&
(
*
it).second
<=
next1.time)
189
continue
;
190
m1[s]
=
next1.time ;
191
qu1.push(next1);
192
}
193
}
194
}
195
}
196
}
197
if
(mark)
198
printf(
"
YES\n
"
);
199
else
200
printf(
"
NO\n
"
);
201
}
202
return
0
;
203
}
204
测试数据：
1
4
3
7
4
5
6
2
1
4
3
7
4
5
6
2
//
YES
2
1
8
8
7
8
1
1
1
5
8
7
7
7
2
4
//
YES
6
5
5
7
5
1
7
1
7
7
6
5
5
7
6
3
//
YES
4
6
5
6
5
5
5
2
2
2
7
6
5
3
5
2
//
YES
4
4
5
5
3
4
7
2
2
7
7
2
6
5
4
8
//
YES
4
2
3
4
5
5
6
6
2
7
4
4
3
5
4
5
//
YES
3
3
5
3
6
2
6
6
6
3
8
3
6
6
6
8
//
YES
3
2
5
3
6
2
6
6
6
3
8
3
6
6
6
8
//
NO
3
4
4
5
5
5
7
5
3
3
4
3
5
3
6
3
//
YES