Stars（一定要看，树状数组差点问线问题）

- 日理万妓 2022-08-01 06:12 194阅读 0赞

# `#include<cstdio> #include<iostream> #include<cstring> const int N=16000; const int Mx=32100; using namespace std; int tree[Mx]; int ans[16000]; int lowbit(int n) { return n&(-n); } int add(int i,int data) { while(i<=Mx) { tree[i]+=data; i+=lowbit(i); } return 0; } int sum(int i) { int sum1=0; while(i>0) { sum1+=tree[i]; i-=lowbit(i); } return sum1; } int main() { int n,x,y,i,m; while(cin>>n) { m=n; memset(ans,0,sizeof(ans)); memset(tree,0,sizeof(tree)); while(n--) { cin>>x>>y; x+=1;//从0开始 add(x,1);//加一 ans[sum(x)-1]++; } for(i=0;i<m;i++) { cout<<ans[i]<<endl; } } }`     Stars #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 5704    Accepted Submission(s): 2260**

Problem 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.    
  
![1541-1.jpg][]  
  
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

Source

[Ural Collegiate Programming Contest 1999][]

**我靠这一题真的很迷糊人，单点更新没有错！**

**这一题几个迷糊人的东西？**

**1.X坐标问题：X坐标怎么实现(1,1)加1之后，当运算到5的时候**

**就可以知道5的点值是2了呢？一直运算啊！lowbit就是算这个的**

**然后算出sum(5)就ok了**

**2.y坐标是干什么的？y坐标其实没用，他们其实都是在更新树状**

**数组的值，比如更新完(1,1)==1之后更新(5,1)==2,那么再更新**

**(3,3)，(3,3)的值是2，（因为他前面的(1,1)更新了）但是此时**

**(5,1)值是3而不是2了，不过没关系！因为(5,1)的值已经计算过**

**了，估计y的作用也就是在此了！**

**发一下代码，注意从1开始更新，既然题目从0开始,x先加1算出**

**sum(x)之后再减1**

#include<cstdio>
    #include<iostream>
    #include<cstring>
    const int N=16000;
    const int Mx=32100;
    using namespace std;
    int tree[Mx];
    int ans[16000];
    int lowbit(int n)
    {
        return n&(-n);
    }
    int add(int i,int data)
    {
        while(i<=Mx)
        {
            tree[i]+=data;
            i+=lowbit(i);
        }
        return 0;
    }
    int sum(int i)
    {
        int sum1=0;
        while(i>0)
        {
            sum1+=tree[i];
            i-=lowbit(i);
        }
        return sum1;
    }
    
    int main()
    {
        int n,x,y,i,m;
        while(cin>>n)
        {
            m=n;
            memset(ans,0,sizeof(ans));
            memset(tree,0,sizeof(tree));
            while(n--)
            {
                cin>>x>>y;
                x+=1;//从0开始
                add(x,1);//加一
                ans[sum(x)-1]++;
            }
            for(i=0;i<m;i++)
            {
                cout<<ans[i]<<endl;
            }
        }
    }

[1541-1.jpg]: /images/20220731/c8c59017af6e463a878a69c4cc2744ed.png
[Ural Collegiate Programming Contest 1999]: http://acm.hdu.edu.cn/search.php?field=problem&key=Ural+Collegiate+Programming+Contest+1999&source=1&searchmode=source