树状数组——HDU2352-优快云博客

本文介绍了一个天文学中关于星图级别的统计问题。通过使用一种特殊的数据结构，文章提供了一种有效的算法来解决如何计算星图上各星的级别，并给出了完整的代码实现。

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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

经典区间和RMQ问题：有n个星星节点，存在星星节点左下角（包括正左和正下）的其他星星节点，则该星星节点比它左下角的星星节点大，level 0表示该星星节点没有比他还小的节点，level 1表示存在一个比该星星节点小的点。输出统计好的每个level等级存在多少星星节点 -来源HDU2352

#include"stdio.h"  
#include"string.h"  
#define N 33000  
int a[N];  
int bit(int x)  
{  
    return x&(-x);  
}  
int sum(int x)  
{  
    int i;  
    int ans=0;  
    for(i=x;i>0;i-=bit(i))  
        ans+=a[i];  
    return ans;  
}  
void add(int x)  
{  
    int i;  
    for(i=x;i<N;i+=bit(i))  
        a[i]++;  
}  
int main()  
{  
    int i;  
    int n;  
    int A[N/2];  
    while(scanf("%d",&n)!=-1)  
    {  
        memset(a,0,sizeof(a));  
        memset(A,0,sizeof(A));  
        int x,y;  
        for(i=0;i<n;i++)  
        {  
            scanf("%d%d",&x,&y);  
            x++;  
            A[sum(x)]++;  
            add(x);  
        }  
        for(i=0;i<n;i++)  
            printf("%d\n",A[i]);  
    }  
    return 0;  
}