poj 2352 stars_线段树基础
Stars
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16362 | Accepted: 7064 |
Description
Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars.
For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.
You are to write a program that will count the amounts of the stars of each level on a given map.
For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.
You are to write a program that will count the amounts of the stars of each level on a given map.
Input
The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.
Output
The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.
Sample Input
5 1 1 5 1 7 1 3 3 5 5
Sample Output
1 2 1 1 0
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
Source
注：这道题目用树状数组是非常方便的，直接进行统计，这里用线段树主要是加强对线段树的理解。这里可以不用lazy思想而直接进行统计。
1
type
2
ji
=^
rec;
3
rec
=
record
4
sum:longint;
5
l,r:longint;
6
lson,rson:ji;
7
end;
8
9
var
10
a:ji;
11
ans:array[
0
..
50005
] of longint;
12
i,j,k,n:longint;
13
procedure build(var a:ji; l,r:longint);
14
var
15
mid:longint;
16
begin
17
new
(a);
18
a
^
.sum:
=
0
;
19
a
^
.l:
=
l;
20
a
^
.r:
=
r;
21
if
l
=
r then exit;
22
mid:
=
(l
+
r)
>>
1
;
23
build(a
^
.lson,l,mid);
24
build(a
^
.rson,mid
+
1
,r);
25
end;
26
27
function sum(x:longint; a:ji):longint;
28
var
29
mid,k:longint;
30
begin
31
k:
=
0
;
32
if
(
1
<=
a
^
.l)and(x
>=
a
^
.r) then
33
begin
34
inc(k,a
^
.sum);
35
exit(k);
36
end;
37
mid:
=
(a
^
.l
+
a
^
.r)
>>
1
;
38
if
1
<=
mid then inc(k,sum(x,a
^
.lson));
39
if
x
>
mid then inc(k,sum(x,a
^
.rson));
40
exit(k);
41
end;
42
43
procedure insert(x:longint; a:ji);
44
var
45
mid:longint;
46
begin
47
if
(x
=
a
^
.l)and(x
=
a
^
.r) then
48
begin
49
inc(a
^
.sum);
50
exit;
51
end;
52
mid:
=
(a
^
.l
+
a
^
.r)
>>
1
;
53
if
x
<=
mid then insert(x,a
^
.lson);
54
if
x
>
mid then insert(x,a
^
.rson);
55
a
^
.sum:
=
a
^
.lson
^
.sum
+
a
^
.rson
^
.sum;
56
end;
57
58
begin
59
readln(n);
60
build(a,
1
,
32001
);
61
for
i:
=
1
to n
do
62
begin
63
readln(j,k);
64
inc(j);
65
inc(ans[sum(j,a)]);
66
insert(j,a);
67
end;
68
for
i:
=
0
to n
-
1
do
writeln(ans[i]);
69
end.