hdu-1541-树状数组-java

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

51 15 17 13 35 5

 

Sample Output

12110

解题思路：星星的y坐标已经按升序排好 题意是输入星星的坐标看他的左下角有多少个星星那么这颗星星的等级就是多少

                    问每个等级（0——n-1）都有多少颗星星 

                    那么也就是 输入xy每次询问 0——x有多少颗星星 也就是区间求和 ，单点更新

                    区间求和 无外乎 差分，线段树，树状数组，

                    本题就是树状数组的经典题

import java.util.Scanner;


public class Main {


private static int n;
private static int[] c;
private static int[] garde;


public static void main(String[] args) {
           Scanner scanner = new Scanner(System.in);
           while (scanner.hasNext()) {
n = scanner.nextInt();
            c = new int [33000];
            garde = new int [n];
            for (int i = 0; i < n; i++) {
int x = scanner.nextInt();
int y = scanner.nextInt();
       garde[sum(x+1)]++;
       add(x+1);
}
            for (int i = 0; i < garde.length; i++) {
System.out.println(garde[i]);
}
}
}


private static void add(int x) {
            while (x<33000) {
c[x]++;
x+=lowbit(x);
}
}


private static int sum(int x) {
int sum = 0;
while (x>0) {
sum+=c[x];
x -= lowbit(x);
}
return sum;
}


private static int lowbit(int x) {
// TODO Auto-generated method stub
return x&(-x);
}


}