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Stars

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.

5
1 1
5 1
7 1
3 3
5 5

1
2
1
1
0

题目：

HDOJ:    http://acm.hdu.edu.cn/showproblem.php?pid=1541

POJ:        http://poj.org/problem?id=2352

又是一道树状数组，要注意题目求的是 每个等级星星的个数。

并不是第i个星星的等级。

等级判别：只要该星星左下方（包括正下方和正左方，不包括自己）有几个星星，等级就是几。

用树状数组的原理，三个函数一上，就出来了。

因为坐标是从0开始，所以相应的坐标要+1操作，否则会崩溃的。

/*******************************************
********************************************
*           Author:Tree                    *
*    From :  blog.csdn.net/lttree          *
*      Title : Stars                       *
*    Source: hdu 1541                      *
*      Hint :  树状数组                    *
********************************************
********************************************/

#include <stdio.h>
#include <string.h>
#define RANGE 32005

// star存等级数，c存星星
int c[RANGE],star[15001];

int lowbit( int x )
{
    return x&(-x);
}
void add( int i )
{
    while( i<=RANGE )
    {
        ++c[i];
        i+=lowbit(i);
    }
}
int sum( int i )
{
    int sum=0;
    while( i>0 )
    {
        sum+=c[i];
        i-=lowbit(i);
    }
    return sum;
}
int main()
{
    int n,i,x,y;
    while( ~scanf("%d",&n) )
    {
        memset(star,0,sizeof(star));
        memset(c,0,sizeof(c));

        for ( i=0;i<n;i++ )
        {
            scanf("%d%d",&x,&y);
            // 坐标是从0开始的，所以x+1
            star[sum(x+1)]++;
            // 后执行add函数，否则会把当前的星星算进去
            add(x+1);
        }
        for( i=0;i<n;i++ )
            printf("%d\n",star[i]);
    }
    return 0;
}

ACM-树状数组之Stars——hdu1541,poj2352,布布扣,bubuko.com

ACM-树状数组之Stars——hdu1541,poj2352

原文：http://blog.csdn.net/lttree/article/details/28616789