POJ 2352 线段树
2012-10-04 14:12
183 查看
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.
![](http://qp.qq.com/cgi-bin/cgi_imgproxy?url=http%3A%2F%2Fpoj.org%2Fimages%2F2352_1.jpg&size=0)
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 g.iven 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
Source::
#include<cstdio>
using namespace std;
#define M 70005
struct Node
{
int l,r;
int n;
}a[M];
int ans[M];
void build(int l,int r,int num)
{
a[num].l=l;
a[num].r=r;
a[num].n=0;
if(l<r)
{
build(l,(l+r)/2,num*2);
build((l+r)/2+1,r,num*2+1);
}
}
void insert(int l,int num)
{
if(a[num].l==l&&a[num].r==l)
{
a[num].n++;
return ;
}
int mid=(a[num].l+a[num].r)/2;
if(l<=mid)
{
insert(l,num*2);
}
else
{
insert(l,num*2+1);
}
a[num].n++;
}
int find(int l,int r,int num)
{
if(a[num].l==l&&a[num].r==r)
{
return a[num].n;
}
int mid=(a[num].l+a[num].r)/2;
if(l>mid)
{
return find(l,r,num*2+1);
}
else if(r<=mid)
{
return find(l,r,num*2);
}
else
{
return find(l,mid,num*2)+find(mid+1,r,num*2+1);
}
}
int main()
{
int x,y,n;
int max=32001;
while(~scanf("%d",&n))
{
build(0,max,1);
for(int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
ans[find(0,x,1)]++;
insert(x,1);
}
for(int i=0;i<n;i++)
{
printf("%d\n",ans[i]);
}
}
return 0;
}
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.
![](http://qp.qq.com/cgi-bin/cgi_imgproxy?url=http%3A%2F%2Fpoj.org%2Fimages%2F2352_1.jpg&size=0)
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 g.iven 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
Source::
#include<cstdio>
using namespace std;
#define M 70005
struct Node
{
int l,r;
int n;
}a[M];
int ans[M];
void build(int l,int r,int num)
{
a[num].l=l;
a[num].r=r;
a[num].n=0;
if(l<r)
{
build(l,(l+r)/2,num*2);
build((l+r)/2+1,r,num*2+1);
}
}
void insert(int l,int num)
{
if(a[num].l==l&&a[num].r==l)
{
a[num].n++;
return ;
}
int mid=(a[num].l+a[num].r)/2;
if(l<=mid)
{
insert(l,num*2);
}
else
{
insert(l,num*2+1);
}
a[num].n++;
}
int find(int l,int r,int num)
{
if(a[num].l==l&&a[num].r==r)
{
return a[num].n;
}
int mid=(a[num].l+a[num].r)/2;
if(l>mid)
{
return find(l,r,num*2+1);
}
else if(r<=mid)
{
return find(l,r,num*2);
}
else
{
return find(l,mid,num*2)+find(mid+1,r,num*2+1);
}
}
int main()
{
int x,y,n;
int max=32001;
while(~scanf("%d",&n))
{
build(0,max,1);
for(int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
ans[find(0,x,1)]++;
insert(x,1);
}
for(int i=0;i<n;i++)
{
printf("%d\n",ans[i]);
}
}
return 0;
}
相关文章推荐
- !POJ 2352 左下角星星-线段树-(单点更新,区间查询)
- POJ 2352 Stars 线段树
- POJ2352 Interval Tree, Segment Tree, 线段树, 区间树
- poj [2352]线段树 Stars
- POJ 2352 Stars Treap & 线段树
- poj 2352 stars_线段树基础
- poj 2352 Stars(线段树)
- poj 2352(入门线段树)
- POJ 2352-Stars(线段树)
- poj_2352 线段树
- POJ 2352 Stars 线段树 单点更新 成段求和
- poj2352-------------线段树------------(*)
- poj 2352 stars 线段树
- POJ 2352 Stars (线段树)
- POJ 2352 Stars 线段树
- poj 2352 Stars(线段树||树状数组)
- 线段树 POJ 2352
- Poj2352-树状数组或线段树
- POJ 2352 Stars (线段树)
- POJ 2352 Stars(树状数组||线段树单点更新)