poj2352 stars

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. 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.

首先根据横坐标排序，这样保证后遍历到的点的横坐标一定大于等于之前的点。

然后只要找出来在之前遍历过的点中有多少纵坐标小于等于这个点。

于是可以用平衡树。

每次查找完之后把这个点插入即可。

#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
struct point
{
int x,y;
bool operator < (const point &a) const
{
return x<a.x||(x==a.x&&y<a.y);
}
}a[100010];
struct node
{
int x,s,v,n,c[2];
}t[100010];
int r,s,ans[100010];
void up(int p)
{
t[p].s=t[t[p].c[0]].s+t[t[p].c[1]].s+t[p].n;
}
void rot(int &p,bool b)
{
int x=t[p].c[b];
int y=t[x].c[!b];
t[x].c[!b]=p;
t[p].c[b]=y;
up(p);
up(x);
p=x;
}
int que(int p,int x)
{
if (!p) return 0;
if (t[p].x==x) return t[p].n+t[t[p].c[0]].s;
if (x<t[p].x) return que(t[p].c[0],x);
return t[t[p].c[0]].s+t[p].n+que(t[p].c[1],x);
}
void ins(int &p,int x)
{
if (p==0)
{
p=++s;
t[p].x=x;
t[p].v=rand();
t[p].n=t[p].s=1;
return;
}
if (x==t[p].x)
{
t[p].n++;
up(p);
return;
}
bool b=x>t[p].x;
ins(t[p].c[b],x);
up(p);
if (t[t[p].c[b]].v>t[p].v)
rot(p,b);
}
int main()
{
int i,j,k,m,n,p,q,x,y,z;
scanf("%d",&n);
for (i=1;i<=n;i++)
scanf("%d%d",&a[i].x,&a[i].y);
sort(a+1,a+n+1);
for (i=1;i<=n;i++)
{
ans[que(r,a[i].y)]++;
ins(r,a[i].y);
}
for (i=0;i<n;i++)
printf("%d\n",ans[i]);
}