hdu 3440 House Man
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 320 Accepted Submission(s): 99
这是本人第一次做差分约束系统的题，以前只是听说，当却不知道什么叫做差分约束系统，现在终于明白了，所谓差分约束系统就是求一组不等式的解。在看本题的约束条件，super man jump<=D 任意相邻两座房子的距离>=1，所意本体就是求满足这些不等式的从最低房子到最高房子的最远距离
代码：
1
#include
<
stdio.h
>
2
#include
<
stdlib.h
>
3
#define
inf 2000000000
4
struct
node
5
{
6
int
h,num;
7
}s[
1005
];
8
struct
bb
9
{
10
int
start,end,z;
11
}bra[
2000
];
12
int
com(
const
void
*
e1,
const
void
*
e2)
13
{
14
struct
node
*
c
=
(node
*
)e1;
15
struct
node
*
d
=
(node
*
)e2;
16
return
c
->
h
-
d
->
h;
17
};
18
int
dist[
1005
],n,k;
19
int
relax(
int
u,
int
v,
int
c)
20
{
21
if
(dist[u]
!=
inf
&&
dist[v]
>
dist[u]
+
c)
22
{
23
dist[v]
=
dist[u]
+
c;
24
return
1
;
25
}
26
return
0
;
27
}
28
int
bellman(
int
v0)
29
{
30
int
i,j;
31
for
(i
=
1
;i
<=
n;i
++
)
32
{
33
dist[i]
=
inf;
34
}
35
dist[v0]
=
0
;
36
int
flag;
37
for
(i
=
1
;i
<
n;i
++
)
38
{
39
flag
=
0
;
40
for
(j
=
1
;j
<=
k;j
++
)
41
{
42
if
(relax(bra[j].start,bra[j].end,bra[j].z))
43
flag
=
1
;
44
}
45
if
(
!
flag)
46
break
;
47
}
48
for
(j
=
1
;j
<=
k;j
++
)
49
{
50
if
(relax(bra[j].start,bra[j].end,bra[j].z))
51
return
0
;
52
}
53
return
1
;
54
}
55
int
max(
int
x,
int
y)
56
{
57
if
(x
>
y)
58
return
x;
59
return
y;
60
}
61
int
min1(
int
x,
int
y)
62
{
63
if
(x
<
y)
64
return
x;
65
return
y;
66
}
67
int
main()
68
{
69
int
t,d,i,a,end,min,max1,v0,j,count
=
0
;
70
scanf(
"
%d
"
,
&
t);
71
while
(t
--
)
72
{
73
min
=
inf;
74
max1
=
0
;k
=
0
;count
++
;
75
scanf(
"
%d%d
"
,
&
n,
&
d);
76
for
(i
=
1
;i
<=
n;i
++
)
77
{
78
scanf(
"
%d
"
,
&
a);
79
s[i].num
=
i;
80
s[i].h
=
a;
81
if
(a
<
min)
82
{
83
v0
=
i;
84
min
=
a;
85
}
86
if
(a
>
max1)
87
{
88
end
=
i;
89
max1
=
a;
90
}
91
}
92
if
(v0
>
end)
93
{
94
int
term
=
v0;
95
v0
=
end;
96
end
=
term;
97
}
98
qsort(s
+
1
,n,
sizeof
(s[
1
]),com);
99
for
(i
=
1
;i
<
n;i
++
)
100
{
101
k
++
;
102
bra[k].start
=
min1(s[i].num,s[i
+
1
].num);
103
bra[k].end
=
max(s[i].num,s[i
+
1
].num);
104
bra[k].z
=
d;
105
k
++
;
106
bra[k].start
=
i
+
1
;
107
bra[k].end
=
i;
108
bra[k].z
=-
1
;
109
}
110
if
(bellman(v0))
111
printf(
"
Case %d: %d\n
"
,count,dist[end]);
112
else
113
printf(
"
Case %d: %d\n
"
,count,
-
1
);
114
}
115
return
0
;
116
}
117