POJ 2352 stars（树状数组）

Stars

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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.

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

刚开始以为用二维的树状数组求矩阵内元素和..然而只会一维的...然后学长提示了这题的y是递增的，也就是说star i+1肯定在star i的右边，因此只需判断是否在下方即可。于是题目变成了一维树状数组的前n项和，若star出现过此X坐标点标记为1即可。然后题目还有个坑点就是X可能是0，然而树状数组要从1开始。因此将题目中每个点均后移1个单位，答案不影响

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
int tree[32010],n;
int pos[15010];
inline int lowbit(const int &k)
{
	return k&-k;
}
inline void add(int k,const int &val)
{
	while (k<=32009)
	{
		tree[k]+=val;
		k+=lowbit(k);
	}
	return ;
}
inline int getsum(int k)
{
	int sum=0;
	while (k)
	{
		sum+=tree[k];
		k-=lowbit(k);
	}
	return sum;
}
int main (void)
{
	ios::sync_with_stdio(false);
	int i,j,val,x,y;
	while (cin>>n)
	{		
		memset(pos,0,sizeof(pos));
		memset(tree,0,sizeof(tree));
		for (i=1; i<=n; i++)
		{
			cin>>x>>y;
			pos[getsum(++x)]++;
			add(x,1);
		}
		for (i=0; i<n; i++)
		{
			cout<<pos[i]<<endl;
		}
	}
	return 0;
}