POJ 1201 Intervals
解题思路：差分约束+spfa
1)dist[e.y]>=dist[e.x]+dist[e.v]
2)dist[i]>=dist[i+1]-1
3)dist[i+1]>=dist[i]
分别将2，3中连续点进行合并Line 56~63
代码
1
#include
<
iostream
>
2
using
namespace
std;
3
#define
MAXN 50005
4
int
n, m, maxX,minX,first[MAXN], next[MAXN
*
3
];
5
struct
{
6
int
y, v;
7
}edge[MAXN
*
3
];
8
inline
void
insert(
int
x,
int
y,
int
v)
9
{
10
edge[m].y
=
y, edge[m].v
=
v;
11
if
(first[x])next[m]
=
first[x];
12
first[x]
=
m
++
;
13
}
14
int
spfa()
15
{
16
int
i,x, y, v,p, s, e,dist[MAXN],que[MAXN];
17
bool
used[MAXN]
=
{
0
};
18
s
=
e
=
0
;
19
for
(i
=
minX;i
<=
maxX;i
++
)dist[i]
=-
MAXN;
20
que[e
++
]
=
minX, dist[minX]
=
0
,used[minX]
=
true
;
21
while
(e
!=
s)
22
{
23
x
=
que[s],p
=
first[x];
24
while
(p)
25
{
26
y
=
edge[p].y, v
=
edge[p].v;
27
if
(dist[y]
<
dist[x]
+
v)
28
{
29
dist[y]
=
dist[x]
+
v;
30
if
(
!
used[y]){used[y]
=
true
;que[e]
=
y,e
=
(e
+
1
)
%
MAXN;}
31
}
32
p
=
next[p];
33
}
34
s
=
(s
+
1
)
%
MAXN;
35
used[x]
=
false
;
36
}
37
return
dist[maxX];
38
}
39
int
main()
40
{
41
int
i, p, q, x, y, v, l;
42
bool
used[MAXN]
=
{
0
};
43
scanf(
"
%d
"
,
&
n);
44
memset(first,
0
,
sizeof
(first));
45
memset(next,
0
,
sizeof
(next));
46
minX
=
MAXN, maxX
=
0
;
47
for
(m
=
i
=
1
;i
<=
n;i
++
)
48
{
49
scanf(
"
%d %d %d
"
,
&
x,
&
y,
&
v);
50
used[x]
=
used[y
+
1
]
=
true
;
51
if
(x
<
minX)minX
=
x;
52
if
(maxX
<
y
+
1
)maxX
=
y
+
1
;
53
insert(x,y
+
1
,v);
54
}
55
p
=
minX;
56
while
(
1
)
57
{
58
q
=
p
+
1
;
59
while
(
!
used[q]
&&
q
<=
maxX)q
++
;
60
if
(q
<=
maxX)insert(p, q,
0
), insert(q, p, p
-
q);
61
else
break
;
62
p
=
q;
63
}
64
l
=
spfa();
65
printf(
"
%d\n
"
, l);
66
return
0
;
67
}