差分约束系统
又学了一遍差分约束系统
SPOJ 1451
NOI 1999的原题，
差分约束SPFA
1
/*
2
* =====================================================================================
3
*
4
* Filename: SPFA.cpp
5
*
6
* Description: SPFA
7
*
8
* Version: 1.0
9
* Created: 2011年03月01日 18时11分01秒
10
* Revision: none
11
* Compiler: gcc
12
*
13
* Author: ronaflx
14
* Company: hit-acm-group
15
*
16
* =====================================================================================
17
*/
18
#include
<
iostream
>
19
#include
<
cstring
>
20
#include
<
cstdio
>
21
#include
<
queue
>
22
using
namespace
std;
23
const
int
V
=
1010
;
24
const
int
E
=
300001
;
25
const
int
INF
=
10000000
;
26
27
struct
edges
28
{
29
int
v, val;
30
edges
*
next;
31
} pool[E],
*
g[V],
*
pp;
32
int
cnt[V], dist[V], q[V];
33
bool
vst[V];
34
35
void
initialize()
36
{
37
memset(g,
0
,
sizeof
(g));
38
pp
=
pool;
39
}
40
41
void
addedge(
int
a,
int
b,
int
v)
42
{
43
pp
->
v
=
b;
44
pp
->
val
=
v;
45
pp
->
next
=
g[a];
46
g[a]
=
pp
++
;
47
}
48
49
void
SPFA(
int
n)
50
{
51
memset(cnt,
0
,
sizeof
(cnt));
52
memset(vst,
false
,
sizeof
(vst));
53
fill(dist, dist
+
n, INF);
54
dist[n
-
1
]
=
0
;
55
vst[n
-
1
]
=
true
;
56
int
tail
=
0
;
57
q[
0
]
=
n
-
1
;
58
for
(
int
i
=
0
;i
<=
tail;i
++
)
59
{
60
int
beg
=
q[i
%
n];
61
vst[beg]
=
false
;
62
for
(edges
*
i
=
g[beg]; i
!=
NULL; i
=
i
->
next)
63
{
64
int
tmp
=
i
->
v, val
=
i
->
val;
65
if
(dist[tmp]
>
dist[beg]
+
val)
66
{
67
dist[tmp]
=
dist[beg]
+
val;
68
if
(
!
vst[tmp])
69
{
70
vst[tmp]
=
true
;
71
cnt[tmp]
++
;
72
tail
++
;
73
q[tail
%
n]
=
tmp;
74
if
(cnt[tmp]
>
n)
75
{
76
puts(
"
-1
"
);
77
return
;
78
}
79
}
80
}
81
}
82
}
83
for
(
int
i
=
1
;i
<
n
-
1
;i
++
)
84
printf(
"
%d
"
, dist[i]
-
dist[i
-
1
]);
85
puts(
""
);
86
}
87
88
int
main()
89
{
90
int
n ,a0, b0, l0, a1, b1, l1;
91
while
(scanf(
"
%d %d %d %d %d %d %d
"
,
&
n,
&
a0,
&
b0,
&
l0,
&
a1,
&
b1,
&
l1)
==
7
)
92
{
93
initialize();
94
for
(
int
i
=
1
;i
+
l0
-
1
<=
n;i
++
)
95
{
96
addedge(i
-
1
, i
+
l0
-
1
, l0
-
a0);
97
addedge(i
+
l0
-
1
, i
-
1
, b0
-
l0);
98
}
99
for
(
int
i
=
1
;i
+
l1
-
1
<=
n;i
++
)
100
{
101
addedge(i
-
1
, i
+
l1
-
1
, b1);
102
addedge(i
+
l1
-
1
, i
-
1
,
-
a1);
103
}
104
for
(
int
i
=
1
;i
<=
n;i
++
)
105
{
106
addedge(i
-
1
, i,
1
);
107
addedge(i, i
-
1
,
0
);
108
}
109
for
(
int
i
=
0
;i
<=
n;i
++
)
110
{
111
addedge(n
+
1
, i,
0
);
112
}
113
SPFA(n
+
2
);
114
}
115
return
0
;
116
}
得出的结论，SPFA的效率比bellman-ford要高很多，没有负环的时候SPFA效率远远要好。
注意人工定点的原因是要把图联通起来，所以设立的。
如果不设立人工定点，只要把所有的点先都扔到队列里就可以了
然后memset(dist, 0, sizeof(dist));再SPFA就可以求解差分约束了。
其他的内容以后再总结。
今天由于一个Bellman-ford 边忘清空了，耽误了一下午的时间。太水了.
先在NOI的网站上看是单case的，到spoj是多case的结果忘清空边了……我怒！！！！
附着题Bellman-Ford的代码，越想越生气，这一天就过去了，浪费了呀
Bellman-ford
1
/*
2
* =====================================================================================
3
*
4
* Filename: Bellman_ford.cpp
5
*
6
* Description: system of difference constraints
7
*
8
* Version: 1.0
9
* Created: 2011年03月01日 14时14分24秒
10
* Revision: none
11
* Compiler: gcc
12
*
13
* Author: ronaflx
14
* Company: hit-acm-group
15
*
16
* =====================================================================================
17
*/
18
#include
<
iostream
>
19
#include
<
cstring
>
20
#include
<
cstdio
>
21
#include
<
vector
>
22
using
namespace
std;
23
const
int
SIZE
=
10110
;
24
const
int
INF
=
100000000
;
25
vector
<
pair
<
pair
<
int
,
int
>
,
int
>
>
edge;
26
int
n, a0, a1, b0, b1, l0, l1;
27
int
d[SIZE];
28
29
void
Bellman_ford(
int
n,
int
p)
30
{
31
fill(d,d
+
n, INF);
32
d[n
-
1
]
=
0
;
33
for
(
int
i
=
1
;i
<
n;i
++
)
34
{
35
bool
unfind
=
true
;
36
for
(vector
<
pair
<
pair
<
int
,
int
>
,
int
>
>
::iterator i
=
edge.begin(); i
!=
edge.end(); i
++
)
37
{
38
int
v
=
i
->
first.first, u
=
i
->
first.second, val
=
i
->
second;
39
if
(d[u]
>
d[v]
+
val)
40
{
41
unfind
=
false
;
42
d[u]
=
d[v]
+
val;
43
}
44
}
45
if
(unfind)
46
break
;
47
}
48
for
(vector
<
pair
<
pair
<
int
,
int
>
,
int
>
>
::iterator i
=
edge.begin(); i
!=
edge.end(); i
++
)
49
{
50
int
v
=
i
->
first.first, u
=
i
->
first.second, val
=
i
->
second;
51
if
(d[u]
>
d[v]
+
val)
52
{
53
puts(
"
-1
"
);
54
return
;
55
}
56
}
57
for
(
int
i
=
1
;i
<=
p;i
++
)
58
printf(
"
%d
"
, d[i]
-
d[i
-
1
]);
59
puts(
""
);
60
}
61
int
main()
62
{
63
while
(scanf(
"
%d %d %d %d %d %d %d
"
,
&
n,
&
a0,
&
b0,
&
l0,
&
a1,
&
b1,
&
l1)
==
7
)
64
{
65
edge.clear();
66
for
(
int
i
=
1
;i
+
l0
-
1
<=
n;i
++
)
67
{
68
edge.push_back(make_pair(make_pair(i
-
1
, i
+
l0
-
1
), l0
-
a0));
69
edge.push_back(make_pair(make_pair(i
+
l0
-
1
, i
-
1
), b0
-
l0));
70
}
71
for
(
int
i
=
1
;i
+
l1
-
1
<=
n;i
++
)
72
{
73
edge.push_back(make_pair(make_pair(i
-
1
, i
+
l1
-
1
), b1));
74
edge.push_back(make_pair(make_pair(i
+
l1
-
1
, i
-
1
),
-
a1));
75
}
76
for
(
int
i
=
1
;i
<=
n;i
++
)
77
{
78
edge.push_back(make_pair(make_pair(i
-
1
, i),
1
));
79
edge.push_back(make_pair(make_pair(i, i
-
1
),
0
));
80
}
81
for
(
int
i
=
0
;i
<=
n;i
++
)
82
edge.push_back(make_pair(make_pair(n
+
1
, i),
0
));
83
Bellman_ford(n
+
2
, n);
84
}
85
return
0
;
86
}