hdu 1198 Farm Irrigation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1066 Accepted Submission(s): 472
听我一同学说他做这一道题时用的是并查集，不过我觉得这一题用广搜更容易，一个一个的扫描，将所有与这个集合里面元素有关的都放进队列里，看能形成几个集合。
代码：
1
2
//
15ms
3
#include
<
stdio.h
>
4
#include
<
queue
>
5
using
namespace
std;
6
typedef
struct
node
7
{
8
char
data;
9
int
x, y;
10
11
}NODE;
12
int
mark[
60
][
60
];
char
a[
60
][
60
];
13
int
main()
14
{
15
int
z[
11
][
4
][
2
]
=
{{{
-
1
,
0
},{
0
,
-
1
},{
0
,
0
},{
0
,
0
}},
16
{{
0
,
-
1
},{
1
,
0
},{
0
,
0
},{
0
,
0
}},
17
{{
-
1
,
0
},{
0
,
1
},{
0
,
0
},{
0
,
0
}},
18
{{
1
,
0
},{
0
,
1
},{
0
,
0
},{
0
,
0
}},
19
{{
0
,
1
},{
0
,
-
1
},{
0
,
0
},{
0
,
0
}},
20
{{
-
1
,
0
},{
1
,
0
},{
0
,
0
},{
0
,
0
}},
21
{{
-
1
,
0
},{
1
,
0
},{
0
,
-
1
},{
0
,
0
}},
22
{{
-
1
,
0
},{
0
,
1
},{
0
,
-
1
},{
0
,
0
}},
23
{{
1
,
0
},{
-
1
,
0
},{
0
,
1
},{
0
,
0
}},
24
{{
1
,
0
},{
0
,
1
},{
0
,
-
1
},{
0
,
0
}},
25
{{
0
,
1
},{
0
,
-
1
},{
1
,
0
},{
-
1
,
0
}}};
26
NODE cur,next;
27
int
m,n,i,j,count,k,tag,t;
28
while
(scanf(
"
%d%d
"
,
&
m,
&
n)
!=
EOF)
29
{
30
31
if
(m
<
0
||
n
<
0
)
32
break
;
33
count
=
0
;
34
for
(i
=
0
;i
<
m;i
++
)
35
{
36
getchar();
37
for
(j
=
0
;j
<
n;j
++
)
38
{
39
scanf(
"
%c
"
,
&
a[i][j]);
40
mark[i][j]
=
0
;
41
}
42
}
43
queue
<
NODE
>
qu;
44
for
(i
=
0
;i
<
m;i
++
)
45
for
(j
=
0
;j
<
n;j
++
)
46
{
47
if
(
!
mark[i][j])
48
{
49
count
++
;
50
mark[i][j]
=
1
;
51
cur.data
=
a[i][j];
52
cur.x
=
j;
53
cur.y
=
i;
54
qu.push(cur);
55
while
(
!
qu.empty())
56
{
57
cur
=
qu.front();
58
qu.pop();
59
for
(k
=
0
;k
<
4
;k
++
)
60
{
61
next.x
=
cur.x
+
z[cur.data
-
'
A
'
][k][
0
];
62
next.y
=
cur.y
+
z[cur.data
-
'
A
'
][k][
1
];
63
tag
=
0
;
64
if
(next.x
>=
0
&&
next.x
<
n
&&
next.y
>=
0
&&
next.y
<
m
&&!
mark[next.y][next.x])
65
{
66
next.data
=
a[next.y][next.x];
67
for
(t
=
0
;t
<
4
;t
++
)
68
{
69
int
x,y;
70
x
=
next.x
+
z[next.data
-
'
A
'
][t][
0
];
71
y
=
next.y
+
z[next.data
-
'
A
'
][t][
1
];
72
if
(x
==
cur.x
&&
y
==
cur.y)
73
{
74
tag
=
1
;
75
break
;
76
}
77
}
78
if
(tag)
79
{
80
mark[next.y][next.x]
=
1
;
81
qu.push(next);
82
}
83
}
84
}
85
}
86
}
87
}
88
printf(
"
%d\n
"
,count);
89
}
90
return
0
;
91
}
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