（POJ - 2352）Stars

最新推荐文章于 2020-10-27 10:15:17 发布

原创最新推荐文章于 2020-10-27 10:15:17 发布 · 457 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#poj

树状数组专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种使用树状数组高效计算平面地图上星星等级分布的方法。星星等级定义为在其左侧及下方星星的数量。通过特定的输入顺序，算法仅需维护星星的X坐标即可实现快速更新与查询。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

（POJ - 2352）Stars

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 49337 Accepted: 21290

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.
这里写图片描述

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

题目大意：按照y的非递减顺序，如果y相同按照x的非递减顺序给出n课星星的坐标。一颗星星的等级是它左下角包含星星得的个数（包括边界，即正左边和正下方），问0 - n-1等级的星星各有多少颗。

思路：既然题目给定了这个输入顺序，一定是有道理的。这样输入的话，对于每一颗星星只需知道他左边有多少颗，因为从下到上读入对于每一颗星星它的y已经计算过，所以只要对x进行操作就可以了，没读入一颗星星的坐标对于这个等级的星星数量++，并更新前面星星的数量。

#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=32005;
int c[maxn],ans[maxn],n;//树状数组c[i]表示x不大于i 

int lowbit(int x)
{
    return x&(-x);
}

void update(int x,int d)
{
    for(;x<=maxn;x+=lowbit(x)) c[x]+=d;
}

int getsum(int x)
{
    int sum=0;
    for(;x>0;x-=lowbit(x)) sum+=c[x];
    return sum;
}

int main()
{
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        memset(ans,0,sizeof(ans));
        for(int i=0;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            ans[getsum(x+1)]++;//x+1的目的是树状数组下标从1开始不能有0
            update(x+1,1); 
        }
        for(int i=0;i<n;i++) printf("%d\n",ans[i]);
    }
    return 0;
}