Stars 单点修改＋区间查询

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

对于一个二维坐标上的点(x, y)，存在x0 <= x && y0 <= y的点(x0, y0)的个数即为点(x, y)的level，求0－n-1 level点的个数

题目的输入比较特殊，X升序，若X相等则按Y升序输入，这样就不用考虑Y了，直接计算当前点之前横坐标小于等于X的个数，还要注意X = 0的时候，幸好这个我注意到了，看讨论里有几个说无限超时的233，就是因为X = 0，所以这里，对每个X实施++　当然，还可以用线段树做，为了简洁方便还是选择树状数组吧

#include 
#include 
#include 
#define lowbit(x) (x & (-x))

using namespace std;
const int N = 32005;

int a[N], c[N];

void Update(int k)
{
    for(int i = k; i <= N; i += lowbit(i))
        c[i] += 1;
}
int GetSum(int k)
{
    int sum = 0;
    for(int i = k; i > 0; i -= lowbit(i))
        sum += c[i];
    return sum;
}
int main()
{
    int n, x, y;
    memset(a, 0, sizeof(a));
    memset(c, 0, sizeof(c));
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d%d", &x, &y);
        x++;
        a[GetSum(x)]++;
        Update(x);       
    }
    for(int i = 0; i < n; i++)
        printf("%d\n", a[i]);
    return 0;
}