D - Stars

最新推荐文章于 2019-08-01 19:29:35 发布

朱仙令

最新推荐文章于 2019-08-01 19:29:35 发布

阅读量405

点赞数

CC 4.0 BY-SA版权

分类专栏：树状数组文章标签：树状数组

本文链接：https://blog.youkuaiyun.com/Foverve1/article/details/52072458

树状数组专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一种使用树状数组来解决星图中星星等级分布问题的方法。通过输入星星的坐标，程序可以计算出每个星星的等级，即位于该星星左侧及下方的星星数量，并统计各级星星的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

题意：求星星的各个等级的星星数量，在星星正左边，正下边以及左下方有多少颗星星，则此星星就为多少级

解法：树状数组，因为Y轴已经为升序，所以不用考虑，这里题意很关键，即星星等级是根据你前面输入的星星判断的，若后面加入的星星比你前面的某颗星星等级低，那么前面的那颗星星也是不会变等级的，正因为这样才能用树状数组做

<span style="font-size:18px;">#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int c[32010],num[15010];
int lowbit(int x)
{
    return x&(-x);
}
void add(int i)
{
    while(i<=32000)
    {
        c[i]++;//对该位置上的星星数量加一，同时更新相关的星星和
        i+=lowbit(i);
    }
}
int read(int i)
{
    int sum=0;
    while(i)
    {
        sum+=c[i];//求出比他等级小的星星个数，因为是先求和再更新此处星星数量，所以没有关系
        i-=lowbit(i);
    }
    return sum;
}
int main()
{
    int n,i,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        memset(c,0,sizeof(c));
        memset(num,0,sizeof(num));
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            int temp=read(x+1);//求在此处星星左下方的星星数量，确定星星等级
            num[temp]++;//对相应等级星星计数，temp肯定是比n小的，仔细想想
            add(x+1);//更新
        }
        for(i=0;i<n;i++)
            printf("%d\n",num[i]);
    }
    return 0;
}
</span>