路漫漫............
算法的学习没有尽头！前面还有无尽远的路！
前两天做一道题的时候看到划分树，找到小航航大牛的博客看了一下，很神奇的算法，总的来说就是一个逆归并的过程，从大到小的归并排序，建立线段树，记录每个区间中被分到左子树的数据，然后查询的时候就可以用log(n)的时间查询任意区间第K大数了。由于本人缺乏创新性，代码严重抄袭小航航牛，不再赘述！
回头用划分树做了北大的2104和杭电的3473，速度还算可以，这里仅给出3473的代码以供参考！
代码
1
#include
<
stdio.h
>
2
#include
<
string
.h
>
3
#include
<
stdlib.h
>
4
#define
M 100005
5
struct
Node{
6
int
l,r;
7
} Tree[M
*
4
];
8
int
sorted[M],toleft[
20
][M],val[
20
][M];
9
__int64 sum[M],tolsum[
20
][M];
10
__int64 lsum,rsum;
11
int
cmp(
const
void
*
a,
const
void
*
b)
12
{
return
*
(
int
*
)a
-
*
(
int
*
)b; }
13
void
build(
int
l,
int
r,
int
d,
int
idx)
14
{
15
Tree[idx].l
=
l; Tree[idx].r
=
r;
16
if
(l
==
r)
return
;
17
int
i,mid
=
(l
+
r)
>>
1
,midv
=
sorted[mid];
18
int
lsame
=
mid
-
l
+
1
;
19
for
(i
=
l; i
<=
r; i
++
)
if
(val[d][i]
<
midv) lsame
--
;
20
int
lpos
=
l, rpos
=
mid
+
1
, same
=
0
;
21
for
(i
=
l; i
<=
r; i
++
){
22
if
(i
==
l){
23
toleft[d][i]
=
0
; tolsum[d][i]
=
0
;
24
}
25
else
{
26
toleft[d][i]
=
toleft[d][i
-
1
]; tolsum[d][i]
=
tolsum[d][i
-
1
];
27
}
28
if
(val[d][i]
<
midv){
29
toleft[d][i]
++
; val[d
+
1
][lpos
++
]
=
val[d][i];
30
tolsum[d][i]
+=
val[d][i];
31
}
32
else
if
(val[d][i]
>
midv) val[d
+
1
][rpos
++
]
=
val[d][i];
33
else
{
34
if
(same
<
lsame){
35
same
++
; toleft[d][i]
++
; val[d
+
1
][lpos
++
]
=
val[d][i];
36
tolsum[d][i]
+=
val[d][i];
37
}
38
else
val[d
+
1
][rpos
++
]
=
val[d][i];
39
}
40
}
41
build(l,mid,d
+
1
,idx
<<
1
);
42
build(mid
+
1
,r,d
+
1
,(idx
<<
1
)
+
1
);
43
}
44
int
query(
int
l,
int
r,
int
k,
int
d,
int
idx)
45
{
46
if
(l
==
r)
return
val[d][l];
47
int
tol;
48
int
stol;
49
__int64 cntl;
50
if
(l
==
Tree[idx].l){
51
stol
=
0
; tol
=
toleft[d][r];
52
cntl
=
0
;
53
}
else
{
54
stol
=
toleft[d][l
-
1
]; tol
=
toleft[d][r]
-
stol;
55
cntl
=
tolsum[d][l
-
1
];
56
}
57
if
(tol
>=
k){
58
int
start
=
Tree[idx].l
+
stol, end
=
start
+
tol
-
1
;
59
60
return
query(start,end,k,d
+
1
,idx
<<
1
);
61
}
else
{
62
lsum
+=
tolsum[d][r]
-
cntl;
63
int
mid
=
(Tree[idx].l
+
Tree[idx].r)
>>
1
;
64
int
tor
=
r
-
l
-
tol
+
1
;
65
int
stor
=
l
-
Tree[idx].l
-
stol;
66
return
query(mid
+
1
+
stor,mid
+
stor
+
tor,k
-
tol,d
+
1
,(idx
<<
1
)
+
1
);
67
}
68
}
69
int
main()
70
{
71
int
ncase,cas,i,k,n,m;
72
scanf (
"
%d
"
,
&
ncase);
73
for
(cas
=
1
; cas
<=
ncase; cas
++
){
74
sum[
0
]
=
0
;
75
scanf (
"
%d
"
,
&
n);
76
for
(i
=
1
; i
<=
n; i
++
){
77
scanf (
"
%d
"
,
&
val[
0
][i]);
78
sorted[i]
=
val[
0
][i];
79
sum[i]
=
sum[i
-
1
]
+
sorted[i];
80
}
81
qsort(sorted
+
1
,n,
sizeof
(sorted[
0
]),cmp);
82
build(
1
,n,
0
,
1
);
83
scanf (
"
%d
"
,
&
m);
84
printf (
"
Case #%d:\n
"
,cas);
85
for
(i
=
0
; i
<
m; i
++
){
86
int
s,e;
87
scanf (
"
%d%d
"
,
&
s,
&
e);
88
s
++
; e
++
;
89
int
num
=
e
-
s
+
1
;
90
k
=
num
/
2
;
if
(num
%
2
!=
0
) k
++
;
91
lsum
=
0
;
92
rsum
=
sum[e]
-
sum[s
-
1
];
93
int
ans
=
query(s,e,k,
0
,
1
);
94
rsum
-=
ans; rsum
-=
lsum;
95
__int64 re
=
(__int64)(k
-
1
)
*
(__int64)ans
-
lsum
+
rsum
-
(__int64)(num
-
k)
*
(__int64)ans;
96
printf (
"
%I64d\n
"
,re);
97
}
98
printf (
"
\n
"
);
99
}
100
return
0
;
101
}
dp和联通分量，下一阶段的目标，路漫漫谁与我同行，水长长自有人同舟！