HDU 1541 Stars(树状数组入门应用)

Stars(传送门)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem 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
题目大意：给一副地图，统计各个等级星星的个数。题目给出的坐标是按y,x排好的(可以忽略y)。那么每次我们读入一个数据（星星），先计算它的等级，然后更新它的位置信息。 星星的等级=它左边星星的数量+加上左下方星星的数量 星 星 的 等 级 = 它 左 边 星 星 的 数 量 + 加 上 左 下 方 星 星 的 数 量 。

#include
#include
#define lowbit(i) ((i)&(-i))
using namespace std;
int c[32004],a[32004];//a[]数组用来保存各个等级星星的数量，c[x]表示当前x星的等级
void update(int x, int v)
{
    for(int i = x; i < 32004; i += lowbit(i))
    {
        c[i] += 1;
    }
}
int getSum(int x)
{
    int sum = 0;
    for(int i = x; i > 0; i -= lowbit(i))
    {
        sum += c[i];
    }
    return sum;
}
int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        memset(c, 0, sizeof c);
        memset(a, 0, sizeof a);
        int x, y;
        for(int i = 0; i < n; i++)
        {
            cin>>x>>y;
            int s = getSum(x+1);//得到等级
            a[s]++;//该等级的星星数量+1
            update(x+1,1);//更新位置信息
        }
        for(int i = 0; i < n; i++)
            cout<return 0;
}