[PKU 1915 2243] 搜索之BFS & A*(续)
{
其实 能用BFS和A*解决的搜索问题很多
比如有一个很simple的骑士巡游问题
就可以这么做
本文只是贴个代码 狗尾续貂 扩充一下
以后有类似的问题也会在这篇文章里补充
}
原题 http://acm.pku.edu.cn/JudgeOnline/problem?id=1915
//2243与这个题基本类似 注意读入输出的细节
用裸BFS 双向BFS A*皆可
如果用A*
应该自己试着去构造一些启发函数
多多尝试 总有一个最好的
我用的就是很简单的曼哈顿距离作启发
效果不错 值得推荐
但不否认有更好的启发函数 下面贴代码
裸BFS
BFS
1
const
maxq
=
100000
;
2
maxs
=
300
;
3
dx:
array
[
1
..
8
]
of
longint
=
(
1
,
2
,
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
);
4
dy:
array
[
1
..
8
]
of
longint
=
(
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
,
1
,
2
);
5
var
n,s,i,gx,gy,t,h,tx,ty:longint;
6
x,y,dep:
array
[
1
..maxq]
of
longint;
7
hash:
array
[
0
..maxs
-
1
,
0
..maxs
-
1
]
of
longint;
8
procedure
main;
9
var
i:longint;
10
begin
11
readln(s);
12
readln(x[
1
],y[
1
]);
13
readln(gx,gy);
14
if
(x[
1
]
=
gx)
and
(y[
1
]
=
gy)
15
then
begin
16
writeln(
0
);
17
exit;
18
end
;
19
fillchar(hash,sizeof(hash),
0
);
20
hash[x[
1
],y[
1
]]:
=
1
;
21
t:
=
1
; h:
=
1
; dep[
0
]:
=
0
;
22
while
h
<=
t
do
23
begin
24
for
i:
=
1
to
8
do
25
begin
26
tx:
=
x[h]
+
dx[i];
27
ty:
=
y[h]
+
dy[i];
28
if
(tx
=
gx)
and
(ty
=
gy)
29
then
begin
30
writeln(dep[h]
+
1
);
31
exit;
32
end
;
33
if
(tx
>=
0
)
and
(tx
<=
s
-
1
)
and
(ty
>=
0
)
and
(ty
<=
s
-
1
)
and
(hash[tx,ty]
=
0
)
34
then
begin
35
hash[tx,ty]:
=
1
;
36
inc(t);
37
dep[t]:
=
dep[h]
+
1
;
38
x[t]:
=
tx; y[t]:
=
ty;
39
end
;
40
end
;
41
inc(h);
42
end
;
43
end
;
44
begin
45
assign(input,
'
knight.in
'
); reset(input);
46
assign(output,
'
knight.out
'
); rewrite(output);
47
readln(n);
48
for
i:
=
1
to
n
do
main;
49
close(input); close(output);
50
end
.
51
双向BFS
BFS-双向
1
const
maxq
=
50000
;
2
maxs
=
300
;
3
dx:
array
[
1
..
8
]
of
longint
=
(
1
,
2
,
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
);
4
dy:
array
[
1
..
8
]
of
longint
=
(
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
,
1
,
2
);
5
var
x,y,dep:
array
[
1
..maxq,
1
..
2
]
of
longint;
6
hash,kind:
array
[
0
..maxs
-
1
,
0
..maxs
-
1
]
of
longint;
7
n,s,i,h1,t1,h2,t2,tx,ty:longint;
8
procedure
main;
9
var
i:longint;
10
begin
11
readln(s);
12
readln(x[
1
,
1
],y[
1
,
1
]);
13
readln(x[
1
,
2
],y[
1
,
2
]);
14
if
(x[
1
,
1
]
=
x[
1
,
2
])
and
(y[
1
,
1
]
=
y[
1
,
2
])
15
then
begin
16
writeln(
0
);
17
exit;
18
end
;
19
fillchar(hash,sizeof(hash),
0
);
20
hash[x[
1
,
1
],y[
1
,
1
]]:
=
1
; hash[x[
1
,
2
],y[
1
,
2
]]:
=
1
;
21
kind[x[
1
,
1
],y[
1
,
1
]]:
=
1
; kind[x[
1
,
2
],y[
1
,
2
]]:
=
2
;
22
h1:
=
1
; t1:
=
1
; h2:
=
1
; t2:
=
1
;
23
dep[
1
,
1
]:
=
0
; dep[
1
,
2
]:
=
0
;
24
while
(h1
<=
t1)
and
(h2
<=
t2)
do
25
begin
26
if
dep[t1,
1
]
<
dep[t2,
2
]
27
then
begin
28
for
i:
=
1
to
8
do
29
begin
30
tx:
=
x[h1,
1
]
+
dx[i];
31
ty:
=
y[h1,
1
]
+
dy[i];
32
if
(tx
<
0
)
or
(tx
>
s
-
1
)
or
(ty
<
0
)
or
(ty
>
s
-
1
)
then
continue;
33
if
hash[tx,ty]
<>
0
34
then
begin
35
if
kind[tx,ty]
=
2
36
then
begin
37
writeln(dep[h1,
1
]
+
1
+
dep[hash[tx,ty],
2
]);
38
exit;
39
end
;
40
continue;
41
end
;
42
inc(t1);
43
x[t1,
1
]:
=
tx; y[t1,
1
]:
=
ty;
44
hash[tx,ty]:
=
t1;
45
kind[tx,ty]:
=
1
;
46
dep[t1,
1
]:
=
dep[h1,
1
]
+
1
;
47
end
;
48
inc(h1);
49
end
50
else
begin
51
for
i:
=
1
to
8
do
52
begin
53
tx:
=
x[h2,
2
]
+
dx[i];
54
ty:
=
y[h2,
2
]
+
dy[i];
55
if
(tx
<
0
)
or
(tx
>
s
-
1
)
or
(ty
<
0
)
or
(ty
>
s
-
1
)
then
continue;
56
if
hash[tx,ty]
<>
0
57
then
begin
58
if
kind[tx,ty]
=
1
59
then
begin
60
writeln(dep[h2,
2
]
+
1
+
dep[hash[tx,ty],
1
]);
61
exit;
62
end
;
63
continue;
64
end
;
65
inc(t2);
66
x[t2,
2
]:
=
tx; y[t2,
2
]:
=
ty;
67
hash[tx,ty]:
=
t2;
68
kind[tx,ty]:
=
2
;
69
dep[t2,
2
]:
=
dep[h2,
2
]
+
1
;
70
end
;
71
inc(h2);
72
end
;
73
end
;
74
end
;
75
begin
76
assign(input,
'
knight.in
'
); reset(input);
77
assign(output,
'
knight.out
'
); rewrite(output);
78
readln(n);
79
for
i:
=
1
to
n
do
main;
80
close(input); close(output);
81
end
.
82
A*
A*
1
const
maxq
=
100000
;
2
maxs
=
300
;
3
dx:
array
[
1
..
8
]
of
longint
=
(
1
,
2
,
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
);
4
dy:
array
[
1
..
8
]
of
longint
=
(
2
,
1
,
-
1
,
-
2
,
-
2
,
-
1
,
1
,
2
);
5
var
x,y,dep,h,f:
array
[
1
..maxq]
of
longint;
6
hash:
array
[
0
..maxs
-
1
,
0
..maxs
-
1
]
of
longint;
7
n,s,i,gx,gy,hs,qs,tx,ty:longint;
8
procedure
main;
9
var
i,j,now,temp:longint;
10
begin
11
readln(s);
12
readln(x[
1
],y[
1
]);
13
readln(gx,gy);
14
if
(x[
1
]
=
gx)
and
(y[
1
]
=
gy)
15
then
begin
16
writeln(
0
);
17
exit;
18
end
;
19
fillchar(hash,sizeof(hash),
0
);
20
hash[x[
1
],y[
1
]]:
=
1
;
21
hs:
=
1
; h[
1
]:
=
1
; f[
1
]:
=
(abs(gx
-
x[
1
])
+
abs(gy
-
y[
1
]))
div
3
;
22
qs:
=
1
; dep[
1
]:
=
0
;
23
while
hs
>
0
do
24
begin
25
now:
=
h[
1
];
26
if
(x[now]
=
gx)
and
(y[now]
=
gy)
27
then
begin
28
writeln(dep[now]);
29
exit;
30
end
;
31
h[
1
]:
=
h[hs]; f[
1
]:
=
f[hs];
32
dec(hs);
33
i:
=
1
;
34
while
true
do
35
begin
36
if
i shl
1
>
hs
then
break;
37
if
(i shl
1
+
1
>
hs)
or
(f[i shl
1
]
<
f[i shl
1
+
1
])
38
then
begin
39
if
f[i shl
1
]
<
f[i]
40
then
begin
41
temp:
=
h[i shl
1
]; h[i shl
1
]:
=
h[i]; h[i]:
=
temp;
42
temp:
=
f[i shl
1
]; f[i shl
1
]:
=
f[i]; f[i]:
=
temp;
43
i:
=
i shl
1
;
44
end
45
else
break;
46
end
47
else
begin
48
if
f[i shl
1
+
1
]
<
f[i]
49
then
begin
50
temp:
=
h[i shl
1
+
1
]; h[i shl
1
+
1
]:
=
h[i]; h[i]:
=
temp;
51
temp:
=
f[i shl
1
+
1
]; f[i shl
1
+
1
]:
=
f[i]; f[i]:
=
temp;
52
i:
=
i shl
1
+
1
;
53
end
54
else
break;
55
end
;
56
end
;
57
for
i:
=
1
to
8
do
58
begin
59
tx:
=
x[now]
+
dx[i];
60
ty:
=
y[now]
+
dy[i];
61
if
(tx
<
0
)
or
(tx
>
s
-
1
)
or
(ty
<
0
)
or
(ty
>
s
-
1
)
or
(hash[tx,ty]
>
0
)
and
(dep[hash[tx,ty]]
<=
dep[now]
+
1
)
then
continue;
62
inc(qs);
63
x[qs]:
=
tx; y[qs]:
=
ty;
64
dep[qs]:
=
dep[now]
+
1
;
65
hash[tx,ty]:
=
qs;
66
inc(hs);
67
h[hs]:
=
qs;
68
f[hs]:
=
dep[qs]
+
(abs(gx
-
x[qs])
+
abs(gy
-
y[qs]))
div
3
;
69
j:
=
hs;
70
while
j
>
1
do
71
begin
72
if
f[j shr
1
]
>
f[j]
73
then
begin
74
temp:
=
h[j shr
1
]; h[j shr
1
]:
=
h[j]; h[j]:
=
temp;
75
temp:
=
f[j shr
1
]; f[j shr
1
]:
=
f[j]; f[j]:
=
temp;
76
j:
=
j shr
1
;
77
end
78
else
break;
79
end
;
80
end
;
81
end
;
82
end
;
83
begin
84
assign(input,
'
knight.in
'
); reset(input);
85
assign(output,
'
knight0.out
'
); rewrite(output);
86
readln(n);
87
for
i:
=
1
to
n
do
main;
88
close(input); close(output);
89
end
.
90
(未完待续)