hdu 1026 Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2699 Accepted Submission(s): 813
Special Judge
本题其实并不难，就是搜索，但是最麻烦的估计就是记录路径。我用的是广搜，在每一个节点中记录下他的前驱节点，这样回溯很容易找到路径。
代码：
1
#include
<
stdio.h
>
2
#include
<
ctype.h
>
3
#include
<
queue
>
4
#include
<
stack
>
5
using
namespace
std;
6
int
dir[
4
][
2
]
=
{{
0
,
1
},{
0
,
-
1
},{
1
,
0
},{
-
1
,
0
}};
7
int
n,m;
8
struct
node
9
{
10
int
time;
11
int
x, y;
12
int
prex,prey;
13
char
data;
14
int
fight;
15
bool
operator
<
(
const
node
&
a)
const
16
{
17
return
a.time
<
time;
18
}
19
}s[
105
][
105
];
20
typedef
struct
t
21
{
22
int
x, y;
23
}T;
24
node cur1;T cur2,next2;
25
int
bfs()
26
{
27
priority_queue
<
node
>
qu;
28
int
i,x1,y1;
29
s[
0
][
0
].time
=
0
;
30
qu.push(s[
0
][
0
]);
31
while
(
!
qu.empty ())
32
{
33
cur1
=
qu.top();
34
qu.pop ();
35
if
(cur1.x
==
n
-
1
&&
cur1.y
==
m
-
1
)
36
return
1
;
37
for
(i
=
0
;i
<
4
;i
++
)
38
{
39
x1
=
cur1.x
+
dir[i][
0
];
40
y1
=
cur1.y
+
dir[i][
1
];
41
if
(x1
>=
0
&&
x1
<
n
&&
y1
>=
0
&&
y1
<
m)
42
{
43
if
(s[x1][y1].data
==
'
.
'
&&
s[x1][y1].time
>
cur1.time
+
1
)
44
{
45
s[x1][y1].time
=
cur1.time
+
1
;
46
s[x1][y1].prex
=
cur1.x;
47
s[x1][y1].prey
=
cur1.y;
48
qu.push(s[x1][y1]);
49
}
50
else
if
(isdigit(s[x1][y1].data)
&&
s[x1][y1].time
>
cur1.time
+
s[x1][y1].data
-
'
0
'
)
51
{
52
s[x1][y1].time
=
cur1.time
+
s[x1][y1].data
-
'
0
'
+
1
;
53
s[x1][y1].fight
=
1
;
54
s[x1][y1].prex
=
cur1.x;
55
s[x1][y1].prey
=
cur1.y;
56
qu.push(s[x1][y1]);
57
}
58
}
59
}
60
}
61
62
return
0
;
63
}
64
int
main()
65
{
66
int
i,j;
char
a;
67
while
(scanf(
"
%d%d
"
,
&
n,
&
m)
!=
EOF)
68
{
69
for
(i
=
0
;i
<
n;i
++
)
70
{
71
getchar();
72
for
(j
=
0
;j
<
m;j
++
)
73
{
74
scanf(
"
%c
"
,
&
a);
75
s[i][j].data
=
a;
76
s[i][j].time
=
0xfffffff
;
77
s[i][j].x
=
i;
78
s[i][j].y
=
j;
79
s[i][j].fight
=
0
;
80
}
81
}
82
if
(bfs())
83
{
84
stack
<
T
>
st;
85
cur2.x
=
n
-
1
;
86
cur2.y
=
m
-
1
;
87
st.push(cur2);
88
while
(
1
)
89
{
90
cur2
=
st.top ();
91
if
(cur2.x
==
0
&&
cur2.y
==
0
)
92
break
;
93
next2.x
=
s[cur2.x][cur2.y].prex;
94
next2.y
=
s[cur2.x][cur2.y].prey;
95
st.push(next2);
96
}
97
printf(
"
It takes %d seconds to reach the target position, let me show you the way.\n
"
,s[n
-
1
][m
-
1
].time);
98
st.pop();
99
while
(
!
st.empty ())
100
{
101
cur2
=
st.top ();
102
if
(s[cur2.x][cur2.y].fight
==
1
)
103
{
104
printf(
"
%ds:(%d,%d)->(%d,%d)\n
"
,s[cur2.x][cur2.y].time
-
(s[cur2.x][cur2.y].data
-
'
0
'
),s[cur2.x][cur2.y].prex ,s[cur2.x][cur2.y].prey,cur2.x,cur2.y);
105
for
(i
=
1
;i
<=
s[cur2.x][cur2.y].data
-
'
0
'
;i
++
)
106
printf(
"
%ds:FIGHT AT (%d,%d)\n
"
,s[cur2.x][cur2.y].time
+
i
-
(s[cur2.x][cur2.y].data
-
'
0
'
),cur2.x ,cur2.y);
107
}
108
else
109
printf(
"
%ds:(%d,%d)->(%d,%d)\n
"
,s[cur2.x][cur2.y].time,s[cur2.x][cur2.y].prex ,s[cur2.x][cur2.y].prey,cur2.x,cur2.y);
110
st.pop ();
111
}
112
}
113
else
114
{
115
printf(
"
God please help our poor hero.\n
"
);
116
}
117
printf(
"
FINISH\n
"
);
118
}
119
return
0
;
120
}
121
122
123