poj 2352 Stars （树状数组）

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 37108		Accepted: 16173

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

依然想不通，智商是硬伤，绝望的想哭，真的想哭。

更新于2015年08月04日01:47:18：

好像有点明白了.....详情看注释

/*************************************************************************
    > File Name: code/poj/2352.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月04日 星期二 00时33分52秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=4E4+7;
int a[N];
int t[N];
int x,y;
int n;
int lowbit (int x)
{
    return x&(-x);
}
void update(int x,int c)
{
    int i;
    for ( int i = x ; i <= 32001 ; i = i + lowbit(i))
    {
    t[i] = t[i] + c;
    }
}
int sum( int x)
{
    int i;
    int res = 0;
    for ( int i = x ; i >= 1 ; i = i-lowbit(i))
    {
    res = res + t[i];
    }
  //  cout<<"x:"<<x<<" res:"<<res<<endl;
    return res;
}
int main()
{
    scanf("%d",&n);
    for ( int i = 0 ; i < n ; i ++ )
    {
    scanf("%d %d",&x,&y);
    a[sum(++x)]++;  // ++x是因为坐标是从0 开始，而树状数组的下标一定是从1开始，不然会TLE
            // sum(++x)是求 ++x 的level
            //
            //
    update(x,1);   //向上更新，对于每一个比x大的位置，由于x的存在，那些位置的点的level都会增加1
                   //比如对于星3，因为星1的读入，我向上更新了星2,星3,星4,星5，这时候他们的level都是1
               //之后读入星2，更新了星3和星5，星3的level是2，星5的level也是2，之后由于星4，星5的level又加1，为5.
    }
    for ( int i = 0 ; i < n ; i ++ )
    {
    printf("%d
",a[i]); //数组a存的是level 为i的点有多少个，即为题目所求。
    }
    return 0;
}