Stars

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

按照y的增序给出星星的坐标 如果有n颗星星在一颗星星的左下角 那这颗星星的等级就是n+1，问每个等级有几颗星星

树状数组题目 每次给出坐标时求一次getsum(x)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int bit[32222],ans[15555];
int n;
int lowbit(int x){
    return x&(-x);
}
int getsum(int x){
    int cnt=0;
    while(x>0){
	cnt+=bit[x];
	x-=lowbit(x);
    }
    return cnt;
}
void add(int x,int a){
   while(x<32222){
      bit[x]+=a;
     x+=lowbit(x);
   }
}
int main(){
    int x,y;
    while(scanf("%d",&n)!=EOF){
	memset(ans,0,sizeof(ans));
	memset(bit,0,sizeof(bit));
	for(int i=1;i<=n;i++){
	    scanf("%d %d",&x,&y);
	    x++;
	    ans[getsum(x)]++;
	    add(x,1);
	}
	for(int i=0;i<n;i++)
	    printf("%d\n",ans[i]);
    }
    return 0;
}