pku 3662 Telephone Lines
刚开始一直没能读懂题，后来读懂题了，感觉没什么思路，后来看discuss里面的二分答案,顿时豁然开朗。。。
先把所有的道路排序，二分一条道路，把长度比当前短的赋值为0，把长度比当前大的赋值为1，用最短路判断从1到n的最短路径ans，如果ans>k 再二分比当前
道路长的路径， 否则二分比当前路径短的路径，直到找到一个最短的路径符合条件。。
贴下代码：
1
# include
<
stdio.h
>
2
# include
<
string
.h
>
3
# include
<
stdlib.h
>
4
# define N
1005
5
# define M
10005
6
# define PI
0xfffffff
7
int
adj[N][N],visit[N],low[N],n;
8
struct
node{
9
int
a,b,val;
10
}s[M];
11
int
cmp(
const
void
*
a,
const
void
*
b)
12
{
13
struct
node
*
c
=
(
struct
node
*
)a;
14
struct
node
*
d
=
(
struct
node
*
)b;
15
return
c
->
val
-
d
->
val;
16
}
17
void
dij()
18
{
19
int
index,min,index1,i;
20
index
=
1
;
21
low[
1
]
=
0
;
22
while
(index
!=
n)
23
{
24
visit[index]
=
1
;
25
min
=
PI;
26
for
(i
=
1
;i
<=
n;i
++
)
27
{
28
if
(visit[i]
==
1
)
continue
;
29
if
(adj[index][i]
==-
1
&&
low[i]
==
PI)
continue
;
30
if
(adj[index][i]
!=-
1
)
31
{
32
if
(low[i]
>
low[index]
+
adj[index][i])
33
low[i]
=
low[index]
+
adj[index][i];
34
}
35
if
(low[i]
<
min) {min
=
low[i];index1
=
i;}
36
}
37
if
(min
==
PI)
break
;
38
index
=
index1;
39
}
40
}
41
int
main()
42
{
43
int
i,m,k,a,b,c,right,left,mid,mid1,ans;
44
while
(scanf(
"
%d%d%d
"
,
&
n,
&
m,
&
k)
!=
EOF)
45
{
46
memset(adj,
-
1
,
sizeof
(adj));
47
for
(i
=
1
;i
<=
m;i
++
)
48
{
49
scanf(
"
%d%d%d
"
,
&
a,
&
b,
&
c);
50
s[i].a
=
a;
51
s[i].b
=
b;
52
s[i].val
=
c;
53
adj[a][b]
=
adj[b][a]
=
1
;
54
}
55
qsort(s
+
1
,m,
sizeof
(s[
1
]),cmp);
56
s[
0
].val
=
0
;
57
mid1
=
0
;
58
right
=
m;left
=
0
;
59
ans
=-
1
;
60
while
(right
>=
left)
61
{
62
mid
=
(right
+
left)
/
2
;
63
if
(mid1
<=
mid)
64
{
65
for
(i
=
mid1;i
<=
mid;i
++
)
66
adj[s[i].a][s[i].b]
=
adj[s[i].b][s[i].a]
=
0
;
67
}
68
else
69
{
70
for
(i
=
mid
+
1
;i
<=
mid1;i
++
)
71
adj[s[i].a][s[i].b]
=
adj[s[i].b][s[i].a]
=
1
;
72
}
73
mid1
=
mid;
74
/*
75
for(i=0;i<=mid;i++)
76
adj[s[i].a][s[i].b]=adj[s[i].b][s[i].a]=0;
77
for(i=mid+1;i<=m;i++)
78
adj[s[i].a][s[i].b]=adj[s[i].b][s[i].a]=1;
*/
79
for
(i
=
0
;i
<=
n;i
++
)
80
{
81
low[i]
=
PI;
82
visit[i]
=
0
;
83
}
84
dij();
85
if
(low[n]
>
k) left
=
mid
+
1
;
86
else
{ans
=
mid;right
=
mid
-
1
;}
87
}
88
if
(ans
==-
1
) printf(
"
-1\n
"
);
89
else
printf(
"
%d\n
"
,s[ans].val);
90
}
91
return
0
;
92
}