树状数组 || POJ 2352 Stars

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

┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉┉┉ ∞ ∞ ┉┉┉┉ ∞ ∞ ┉┉┉

题意：给出一堆星星的坐标，求每个星星左侧和下侧空间内的所有星星数量

思路：因为发现输入里y是递增输入的，所以输入当前这个星星的时候，它前面的所有星星肯定都在它下面（或者和它一行）

就转化为求之前输入的那些星星有多少个x坐标比它小的，用树状数组来维护

树状数组。。还不怎么会用哇。。。

数组大小开的。。emm是一个迷

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
#define SZ 320200
int a[SZ], d[SZ], t[SZ], n;
void add(int i, int x)
{
    for(; i <= ; i += (i & -i))
        d[i] += x;
}
int get(int i)
{
    int ans = ;
    for(; i; i -= (i & -i))
        ans += d[i];
    return ans;
}

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = ; i < n; i++)
    {
        int x, y;
        scanf("%d %d", &x, &y);
        x++;//x可能等于0，把它们都+1
        int ans = get(x);//求x之前的所有的和（前缀和，就是当前点的等级
        t[ans]++;
        add(x, );
    }
    for(int i = ; i < n; i++) printf("%d\n", t[i]);
    return ;
}