Usaco2007Jan Balanced Lineup [bzoj 1636]
♦♦♦♦♦♦ Description ♦♦♦♦♦♦
For the daily milking, Farmer John’s N cows (1 <= N <= 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height. Farmer John has made a list of Q (1 <= Q <= 200,000) potential groups of cows and their heights (1 <= height <= 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.每天,农夫 John 的N头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大.
John 准备了Q个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.
注意: 在最大数据上, 输入和输出将占用大部分运行时间.
John 准备了Q个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.
注意: 在最大数据上, 输入和输出将占用大部分运行时间.
♦♦♦♦♦♦ Input ♦♦♦♦♦♦
* Line 1: Two space-separated integers, N and Q. * Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i * Lines N+2..N+Q+1: Two integers A and B (1 <= A <= B <= N), representing the range of cows from A to B inclusive.
♦♦♦♦♦♦ Output ♦♦♦♦♦♦
* Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
♦♦♦♦♦♦ Sample Input ♦♦♦♦♦♦
6 3
1
7
3
4
2
5
1 5
4 6
2 2
♦♦♦♦♦♦ Sample Output ♦♦♦♦♦♦
6
3
0
♦♦♦♦♦♦ Hint ♦♦♦♦♦♦
1 <= N <= 50,000
1 <= Q <= 180,000
♦♦♦♦♦♦ 题解 ♦♦♦♦♦♦
没有修改的区间最大最小值，直接用树状数组做就可以啦
可是翻了一下Status发现只有我一个人用的树状数组似乎，没错那个内存三位数的就是我！不过时间怎么会慢了一点的说….
当然这道题用线段树才是简单易懂，一些无聊的神犇还会套主席树的模板… 果然还是我太弱了啊…
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var
minn
,
maxx
,
a
:
array
[
0..50001
]
of
longint
;
n
,
q
,
i
,
l
,
r
,
x
,
ans1
,
ans2
:
longint
;
function
max
(
a
,
b
:
longint
)
:
longint
;
begin
if
a
>
b
then
exit
(
a
)
else
exit
(
b
)
;
end
;
function
min
(
a
,
b
:
longint
)
:
longint
;
begin
if
a
>
b
then
exit
(
b
)
else
exit
(
a
)
;
end
;
function
lowbit
(
x
:
longint
)
:
longint
;
begin
exit
(
x
and
(-
x
)
)
;
end
;
procedure
ins
(
p
,
w
:
longint
)
;
begin
while
p
<=
n
do
begin
minn
[
p
]
:=
min
(
minn
[
p
]
,
w
)
;
maxx
[
p
]
:=
max
(
maxx
[
p
]
,
w
)
;
inc
(
p
,
lowbit
(
p
)
)
;
end
;
end
;
function
find
(
l
,
r
:
longint
)
:
longint
;
begin
ans1
:=
0
;
ans2
:=
maxlongint
;
while
true
do
begin
ans1
:=
max
(
ans1
,
a
[
r
]
)
;
ans2
:=
min
(
ans2
,
a
[
r
]
)
;
if
l
=
r
then
break
;
dec
(
r
)
;
while
r
-
l
>=
lowbit
(
r
)
do
begin
ans1
:=
max
(
ans1
,
maxx
[
r
]
)
;
ans2
:=
min
(
ans2
,
minn
[
r
]
)
;
dec
(
r
,
lowbit
(
r
)
)
;
end
;
end
;
exit
(
ans1
-
ans2
)
;
end
;
begin
readln
(
n
,
q
)
;
for
i
:=
1
to
n
do
begin
maxx
[
i
]
:=
0
;
minn
[
i
]
:=
maxlongint
;
end
;
for
i
:=
1
to
n
do
begin
readln
(
a
[
i
]
)
;
ins
(
i
,
a
[
i
]
)
;
end
;
for
i
:=
1
to
q
do
begin
readln
(
l
,
r
)
;
writeln
(
find
(
l
,
r
)
)
;
end
;
end
.
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