poj 3026 Borg Maze
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3255 | Accepted: 1044 |
本题其实并不难，就是题意令人太难理解，后来才看出来，题意的意思是从一个出发点‘S’出发寻找‘A’，但是可以并行查找，即在开始的时候就分组，或者是找到一个‘A’以后再分组。求最小的距离。其实就是一个最小生成树问题，只不过加了一个BFS罢了！
代码：
1
#include
<
stdio.h
>
2
#include
<
string
.h
>
3
#include
<
queue
>
4
using
namespace
std;
5
int
m,nn;
char
map[
55
][
55
];
6
typedef
struct
node
7
{
8
int
x,y,s;
9
}NODE;
10
typedef
struct
pnode
11
{
12
int
s,x,y;
13
bool
operator
<
(
const
pnode
&
a )
const
14
{
15
return
a.s
<
s;
16
}
17
}PNODE;
18
NODE cur,next;PNODE pcur;
19
int
bfs(
int
x,
int
y)
20
{
21
priority_queue
<
PNODE
>
pqu;
22
queue
<
NODE
>
qu;
23
int
visit[
55
][
55
];
24
memset(visit,
0
,
sizeof
(visit));
25
int
mark[
55
][
55
],sum
=
0
,i,tag
=
0
;
26
int
dir[
4
][
2
]
=
{{
0
,
1
},{
0
,
-
1
},{
1
,
0
},{
-
1
,
0
}};
27
pcur.x
=
x;pcur.y
=
y;
28
pcur.s
=
0
;
29
pqu.push(pcur);
30
while
(
!
pqu.empty())
31
{
32
pcur
=
pqu.top();
33
while
(visit[pcur.x][pcur.y])
34
{
35
pqu.pop();
36
if
(pqu.empty ())
37
{
38
tag
=
1
;
39
break
;
40
}
41
pcur
=
pqu.top();
42
}
43
if
(tag)
44
return
sum;
45
pqu.pop();
46
visit[pcur.x][pcur.y]
=
1
;
47
sum
+=
pcur.s;
48
cur.x
=
pcur.x;
49
cur.y
=
pcur.y;
50
cur.s
=
0
;
51
qu.push(cur);
52
memset(mark,
0
,
sizeof
(mark));
53
mark[cur.x][cur.y]
=
1
;
54
while
(
!
qu.empty())
55
{
56
cur
=
qu.front();
57
qu.pop();
58
if
(
!
visit[cur.x][cur.y]
&&
map[cur.x][cur.y]
==
'
A
'
)
59
{
60
pcur.x
=
cur.x;
61
pcur.y
=
cur.y;
62
pcur.s
=
cur.s;
63
pqu.push(pcur);
64
}
65
for
(i
=
0
;i
<
4
;i
++
)
66
{
67
next.x
=
cur.x
+
dir[i][
0
];
68
next.y
=
cur.y
+
dir[i][
1
];
69
if
(next.x
>=
1
&&
next.x
<=
m
&&
next.y
>=
1
&&
next.y
<=
nn
&&
!
mark[next.x][next.y]
&&
map[next.x][next.y]
!=
'
#
'
)
70
{
71
mark[next.x][next.y]
=
1
;
72
next.s
=
cur.s
+
1
;
73
qu.push(next);
74
}
75
}
76
}
77
}
78
return
sum;
79
}
80
int
main()
81
{
82
int
n,i,j,sx,sy;
83
scanf(
"
%d
"
,
&
n);
84
while
(n
--
)
85
{
86
scanf(
"
%d%d
"
,
&
nn,
&
m);
87
for
(i
=
1
;i
<=
m;i
++
)
88
{
89
while
(getchar()
!=
'
\n
'
);
90
for
(j
=
1
;j
<=
nn;j
++
)
91
{
92
scanf(
"
%c
"
,
&
map[i][j]);
93
if
(map[i][j]
==
'
S
'
)
94
{
95
sx
=
i;sy
=
j;
96
}
97
}
98
}
99
printf(
"
%d\n
"
,bfs(sx,sy));
100
}
101
return
0
;
102
}