hdu 1142 A Walk Through the Forest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1214 Accepted Submission(s): 394
本题的思想就是dfs+dijkstra。本题的意思是，如果在点B出存在一条路使得从B点出发可以比从A点出发跟快的到达home，所以B到home的最短距离要比A到home得最短距离小，所以本题首先要求出各点到home得距离，然后再用记忆搜索法搜索即可
代码：
1
#include
<
stdio.h
>
2
#include
<
string
.h
>
3
#include
<
queue
>
4
using
namespace
std;
5
int
dist[
1005
],dp[
1005
];
6
typedef
struct
t
7
{
8
int
end,len;
9
struct
t
*
next;
10
}T;
11
struct
node
12
{
13
int
data;
14
int
len;
15
bool
operator
<
(
const
node
&
a)
const
16
{
17
return
a.len
<
len;
18
}
19
T
*
next;
20
}s[
1005
];
21
void
dijkstra(
int
v0)
22
{
23
s[v0].len
=
0
;node cur;
24
int
visit[
1005
],mark
=
0
;
25
memset(visit,
0
,
sizeof
(visit));
26
memset(dist,
-
1
,
sizeof
(dist));
27
priority_queue
<
node
>
qu;
28
qu.push(s[v0]);
29
dist[v0]
=
0
;
30
while
(
!
qu.empty ())
31
{
32
cur
=
qu.top();
33
while
(visit[cur.data])
34
{
35
qu.pop();
36
if
(qu.empty ())
37
{
38
mark
=
1
;
39
break
;
40
}
41
cur
=
qu.top();
42
}
43
if
(mark)
44
break
;
45
qu.pop();
46
visit[cur.data]
=
1
;
47
T
*
p
=
cur.next;
48
while
(p)
49
{
50
if
(
!
visit[p
->
end])
51
{
52
if
(dist[p
->
end]
==-
1
||
dist[cur.data]
+
p
->
len
<
dist[p
->
end] )
53
s[p
->
end].len
=
dist[p
->
end]
=
p
->
len
+
dist[cur.data];
54
qu.push (s[p
->
end]);
55
}
56
p
=
p
->
next;
57
}
58
}
59
}
60
int
dfs(
int
v0)
61
{
62
int
temp
=
0
;
63
if
(v0
==
2
)
64
return
1
;
65
if
(dp[v0]
!=-
1
)
66
return
dp[v0];
67
t
*
p
=
s[v0].next;
68
while
(p)
69
{
70
if
(dist[v0]
>
dist[p
->
end])
71
{
72
dp[p
->
end]
=
dfs(p
->
end);
73
temp
+=
dp[p
->
end];
74
}
75
p
=
p
->
next;
76
}
77
return
temp;
78
}
79
int
main()
80
{
81
int
n,i,m,a,b,len;
82
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
83
{
84
if
(n
==
0
)
85
break
;
86
scanf(
"
%d
"
,
&
m);
87
T
*
p,
*
q;
88
for
(i
=
1
;i
<=
n;i
++
)
89
{
90
s[i].next
=
NULL;
91
s[i].data
=
i;
92
}
93
memset(dp,
-
1
,
sizeof
(dp));
94
for
(i
=
1
;i
<=
m;i
++
)
95
{
96
scanf(
"
%d%d%d
"
,
&
a,
&
b,
&
len);
97
p
=
(T
*
)malloc(
sizeof
(T));
98
p
->
end
=
b;
99
p
->
len
=
len;
100
p
->
next
=
s[a].next;
101
s[a].next
=
p;
102
q
=
(T
*
)malloc(
sizeof
(T));
103
q
->
end
=
a;
104
q
->
len
=
len;
105
q
->
next
=
s[b].next;
106
s[b].next
=
q;
107
}
108
dijkstra(
2
);
109
printf(
"
%d\n
"
,dfs(
1
));
110
}
111
return
0
;
112
}
113