Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7737    Accepted Submission(s): 3070

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

Source

Ural Collegiate Programming Contest 1999

题意:平面上有n颗星星,每颗星星的等级是由其左下角的星星数量决定的，问每个等级的星星有多少?注意：星星的输入是按y轴递增输入的，y相等则按x轴递增输入。

题解：这题是怎么想到树状数组的。。。。。因为y轴是按照递增输入的，所以只要考虑x轴的，每次输入x轴的坐标,我们先计算一下其左边的点的数量,这里用到了求和操作，因为已经按照y轴排好序,所以小于等于x的我们都可以加上，然后我们更新其后面x轴大于它的所有点。统计输出即可。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 32005;
int c[N];
int res[N];
int n;
int lowbit(int x){
    return x&(-x);
}
void update(int idx,int x){
    for(int i=idx;i<=N;i+=lowbit(i)){
        c[i]+=x;
    }
}
int sum(int idx){
    int v=0;
    for(int i=idx;i>=1;i-=lowbit(i)){
        v+=c[i];
    }
    return v;
}
int main()
{

    while(scanf("%d",&n)!=EOF){
        memset(c,0,sizeof(c));
        memset(res,0,sizeof(res));
        for(int i=1;i<=n;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            x++;
            res[sum(x)]++;
            update(x,1);
        }
        for(int i=0;i<n;i++){
            printf("%d\n",res[i]);
        }
    }
    return 0;
}