HDU1541 树状数组

Posted 2020-08-19 LuZhiyuan

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8938    Accepted Submission(s): 3551

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.



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

 

题意：

给出N个星星的坐标，给出顺序是先按Y从小到大，后按X从小到大。问在某颗星星的左下方分别有0,1,2，.....n-1颗星星的星星坐标有几个，例如上面左下方有0课星星的有1号这一个，左下方有1颗星星的有2号和4号这两个。

代码：

 1 //坑爹的题，竟然已经按y,x排好序了,本想用二维树状数组。坑爹这题竟然ac了之后才有思路，首先按y排好序右安x值排好序了这样
 2 //只需要每输入一个算一个就行，因为y相同的不用解释，y不同的后输入的Y大于前面的，X值同样也大于他左边的。
 3 #include<cstdio>
 4 #include<iostream>
 5 #include<cstring>
 6 using namespace std;
 7 int A[32003];
 8 int lowbit(int x)
 9 {
10     return x&(-x);
11 }
12 void add(int idx)
13 {
14     for(int i=idx;i<=32003;i+=lowbit(i))
15     A[i]++;
16 }
17 int sum(int idx)
18 {
19     int s=0;
20     for(int i=idx;i>0;i-=lowbit(i))
21     s+=A[i];
22     return s;
23 }
24 int main()
25 {
26     int n,a,b,c,cnt,t;
27     int ans[32003];
28     while(scanf("%d",&n)!=EOF)
29     {
30         memset(A,0,sizeof(A));
31         memset(ans,0,sizeof(ans));
32         for(int i=1;i<=n;i++)
33         {
34             scanf("%d%d",&a,&b);
35             add(a+1);
36             ans[sum(a+1)]++;
37             
38         }
39         for(int i=1;i<=n;i++)
40         printf("%d\n",ans[i]);
41     }
42     return 0;
43 }