POJ 2352 Stars（樹狀數組）

Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35467   Accepted: 15390

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



題意：問在輸入的點（星星）的坐標的左下方（包含左方和下方）有幾個點（星星）。有幾個點（星星）就是第幾種情況，一共有1~n種情況，輸出每種情況的個數。因為題目保證y的坐標是按升序輸入的，所以就不用考慮y，只需要更新x的樹狀數組就可以了。

#include
#include
#include
#include
#include

using namespace std;

const int maxn = 32001;

int c[maxn];
int v[maxn];
int n;

int lowbit(int x)
{
    return x&(-x);
}

void updata(int i,int k)
{
    while(i<=maxn)
    {
        c[i] += k;
        i += lowbit(i);
    }
}

int getsum(int i)
{
    int sum = 0;
    while(i>0)
    {
        sum += c[i];
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int x,y;
        memset(v,0,sizeof(v));
        memset(c,0,sizeof(c));
        for(int i=0;i