HDU - 1541——star

最新推荐文章于 2020-08-18 11:29:10 发布

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最新推荐文章于 2020-08-18 11:29:10 发布

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分类专栏：线段树

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本文介绍了一个天文学领域的算法问题，即如何通过编程计算星图中每个星星的级别，并统计各个级别的星星数量。级别定义为一个星星左侧下方星星的数量。文章提供了一个具体的C++实现方案，利用线段树进行区间更新和查询。

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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：大概就是计算每颗星星左下边包括了多少颗星星，这个数值就是level。左下边不包括本身，不超过本身的x，y的坐标，可以等于。问level 0-n-1各有多少颗星星。

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 32005
using namespace std;
int sum[N<<2],num[15002];
void update(int rf,int l,int r,int a)
{
    if(l==r)
    {
        sum[rf]++;
        return;
    }
    int mid=(l+r)>>1;
    if(a<=mid)
        update(rf*2,l,mid,a);
    else
        update(rf*2+1,mid+1,r,a);
    sum[rf]=sum[rf*2]+sum[rf*2+1];
}
int query(int rf,int l,int r,int a,int b)
{
    if(a<=l&&b>=r)
    {
        return sum[rf];
    }
    int mid=(l+r)>>1;
    int ans=0;
    if(a<=mid)
        ans+=query(rf*2,l,mid,a,b);
    if(b>mid)
        ans+=query(rf*2+1,mid+1,r,a,b);
    return ans;
}
int main()
{
    int n,y,i,x;
    while(~scanf("%d",&n))
    {
        memset(sum,0,sizeof(sum));
        memset(num,0,sizeof(num));
        for(i=1; i<=n; i++)
        {
            scanf("%d%d",&x,&y);
            update(1,0,N,x);
            num[query(1,0,N,0,x)]++;
        }
        for(i=1; i<=n; i++)
            printf("%d\n",num[i]);
    }
    return 0;
}