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最新推荐文章于 2019-07-25 20:16:38 发布

原创最新推荐文章于 2019-07-25 20:16:38 发布 · 594 阅读

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#insert #build #output #numbers #input #each

本文介绍了一种使用树状数据结构来高效统计二维平面上星星分布级别的算法。该算法通过构建区间树并动态更新星星出现的位置，进而计算每个星星的级别（即不高于且不位于其右侧的星星数量）。文章提供了一个完整的实现示例，展示了如何处理输入数据，构建和维护区间树，并最终输出每个级别的星星数量。

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 g.iven 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::
#include<cstdio>
using namespace std;
#define M 70005
struct Node
{
    int l,r;
    int n;
}a[M];
int ans[M];
void build(int l,int r,int num)
{
    a[num].l=l;
    a[num].r=r;
    a[num].n=0;
    if(l<r)
    {
        build(l,(l+r)/2,num*2);
        build((l+r)/2+1,r,num*2+1);
    }
}

void insert(int l,int num)
{
    if(a[num].l==l&&a[num].r==l)
    {
        a[num].n++;
        return ;
    }
    int mid=(a[num].l+a[num].r)/2;
    if(l<=mid)
    {
        insert(l,num*2);
    }
    else
    {
        insert(l,num*2+1);
    }
    a[num].n++;
}

int find(int l,int r,int num)
{
    if(a[num].l==l&&a[num].r==r)
    {
        return a[num].n;
    }
    int mid=(a[num].l+a[num].r)/2;
    if(l>mid)
    {
        return find(l,r,num*2+1);
    }
    else if(r<=mid)
    {
        return find(l,r,num*2);
    }
    else
    {
        return find(l,mid,num*2)+find(mid+1,r,num*2+1);
    }
}
int main()
{
    int x,y,n;
    int max=32001;
    while(~scanf("%d",&n))
    {
        build(0,max,1);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            ans[find(0,x,1)]++;
            insert(x,1);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}