HDU 3473 Minimum Sum【划分树】
http://acm.hdu.edu.cn/submit.php?pid=3473
HDU 3473 Minimum Sum
大意：已知n个正整数x0,x1,x2...xn-1排成一列，要求回答若干问题，
问题类型如下：
给你一个区间[l,r],要求sum(x-xi)(l<=i<=r)的最小值,其中x必须为xl,xl+1...xr中的一个数
分析：易推得对于任意区间[l,r],符合条件的x为所在区间的中位数，求任意区间的中位数可以用划分树
来解决，同样的，我们用suml[deep][i]来记录划分树中第deep层到数i位置放入左子树的数字的和，
注意：当最后得出中位数median左边的和suml及中位数右边的和sumr,以及它们所对应的个数lnum,rnum,
此时ans = sumr - suml +(lnum-rnum)*ave,切不可左右分开求，否者数据可能超过__int64,我就死在这里了，纠结了整整一下午。。~~~~(>_<)~~~~
View Code
#include
<
stdio.h
>
#include
<
iostream
>
#include
<
string
.h
>
#include
<
algorithm
>
using
namespace
std;
const
int
MAXN
=
100000
+
10
;
typedef __int64 LL;
long
tree[
22
][MAXN],toleft[
22
][MAXN];
LL lsum[
22
][MAXN];
//
lsum[deep][i]为第deep层到第xi被划入左子树的数字的和
LL sum[MAXN];
//
sum[i]为1。。i的和
long
sorted[MAXN];
//
已经排好序的数据
LL suml;
int
lnum;
void
build_tree(
int
left,
int
right,
int
deep)
//
建树
{
if
(left
==
right)
return
;
int
mid
=
(left
+
right)
>>
1
;
int
i;
int
same
=
mid
-
left
+
1
;
//
位于左子树的数据
for
(i
=
left;i
<=
right;i
++
)
//
计算放于左子树中与中位数相等的数字个数
{
if
(tree[deep][i]
<
sorted[mid])
same
--
;
}
int
ls
=
left;
int
rs
=
mid
+
1
;
for
(i
=
left;i
<=
right;i
++
)
{
int
flag
=
0
;
if
((tree[deep][i]
<
sorted[mid])
||
(tree[deep][i]
==
sorted[mid]
&&
same
>
0
))
//
放入左子树
{
flag
=
1
;
tree[deep
+
1
][ls
++
]
=
tree[deep][i];
if
(tree[deep][i]
==
sorted[mid])same
--
;
lsum[deep][i]
=
lsum[deep][i
-
1
]
+
tree[deep][i];
}
else
//
右子树
{
tree[deep
+
1
][rs
++
]
=
tree[deep][i];
lsum[deep][i]
=
lsum[deep][i
-
1
];
}
toleft[deep][i]
=
toleft[deep][i
-
1
]
+
flag;
}
build_tree(left,mid,deep
+
1
);
build_tree(mid
+
1
,right,deep
+
1
);
}
int
query(
int
left,
int
right,
int
k,
int
L,
int
R,
int
deep)
{
if
(left
==
right)
return
tree[deep][left];
int
mid
=
(L
+
R)
>>
1
;
int
x
=
toleft[deep][left
-
1
]
-
toleft[deep][L
-
1
];
//
位于left左边的放于左子树中的数字个数
int
y
=
toleft[deep][right]
-
toleft[deep][L
-
1
];
//
到right为止位于左子树的个数
int
ry
=
right
-
L
-
y;
//
到right右边为止位于右子树的数字个数
int
cnt
=
y
-
x;
//
[left,right]区间内放到左子树中的个数
int
rx
=
left
-
L
-
x;
//
left左边放在右子树中的数字个数
if
(cnt
>=
k)
return
query(L
+
x,L
+
y
-
1
,k,L,mid,deep
+
1
);
else
{
lnum
=
lnum
+
cnt;
suml
=
suml
+
lsum[deep][right]
-
lsum[deep][left
-
1
];
return
query(mid
+
rx
+
1
,mid
+
1
+
ry,k
-
cnt,mid
+
1
,R,deep
+
1
);
}
}
int
main()
{
//
freopen("E:/MYACM/Match/UESTC-4/Minimum Sum/1.in","r",stdin);
//
freopen("2.out","w",stdout);
int
n,m;
int
T;
int
cases
=
0
;
while
(scanf(
"
%d
"
,
&
T)
!=
EOF)
{
cases
=
0
;
while
(T
--
)
{
cases
++
;
scanf(
"
%d
"
,
&
n);
int
i;
for
(i
=
0
;i
<
20
;i
++
)
lsum[i][
0
]
=
0
;
sum[
0
]
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
{
scanf(
"
%d
"
,
&
sorted[i]);
tree[
0
][i]
=
sorted[i];
sum[i]
=
sum[i
-
1
]
+
sorted[i];
}
sort(sorted
+
1
,sorted
+
n
+
1
);
build_tree(
1
,n,
0
);
scanf(
"
%d
"
,
&
m);
printf(
"
Case #%d:\n
"
,cases);
while
(m
--
)
{
int
left,right,k;
scanf(
"
%d%d
"
,
&
left,
&
right);
left
++
;
right
++
;
k
=
((right
-
left)
/
2
)
+
1
;
suml
=
0
;
lnum
=
0
;
long
ave
=
query(left,right,k,
1
,n,
0
);
//
求中位数
int
rnum
=
(right
-
left
+
1
-
lnum);
LL sumr
=
sum[right]
-
sum[left
-
1
]
-
suml;
//
右子树的和
LL ans
=
sumr
-
ave
*
(rnum
-
lnum)
-
suml;
printf(
"
%I64d\n
"
,ans);
}
printf(
"
\n
"
);
}
}
return
0
;
}