POJ 2352 Stars - 树状数组/线段树

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.

此题需要考虑到条件所给的Y升序排列的意义，即按此顺序进行update即可消去y这一维度，避免使用二维树状数组或二维二叉树，最后使用num[]统计个数。
N.B.存在坐标为0的情况，因此读入后需要+1。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;

const int maxn=32100;

int c[maxn],num[maxn];
int n,m;

inline int lowbit(int x)
{
return x&(-x);
}
void update(int x)
{
for(int i=x;i<=maxn;i+=lowbit(i))
c[i]++;
}
int query(int x)
{
int ans=0;
for(int i=x;i>0;i-=lowbit(i))
ans+=c[i];
return ans;
}
int main()
{
scanf("%d",&n);
int x,y;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
num[query(x+1)]++;
update(x+1);
}
for(int i=0;i<n;i++)
printf("%d\n",num[i]);
return 0;
}