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[ACM] POJ2352 [树状数组稍微变形]

2024-07-02 06:30:14   226人阅读

 

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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<memory.h>
 4 using namespace std;
 5 #define maxn 32100//是叶节点能达到的最大值多一点
 6 int C[maxn];//从1开始编号
 7 //--------------------------
 8 int lowbit(int x){
 9     return x&-x;
10 }
11 int sum(int x){
12     int ret=0;
13     while(x>0){
14         ret+=C[x];
15         x-=lowbit(x);
16     }
17     return ret;
18 }
19 void add(int x,int d){
20     while(x<=maxn){
21         C[x]+=d;
22         x+=lowbit(x);
23     }
24 }
25 //--------------------------
26 /*
27 给定n个点，问这一点左下角的区域的点的个数，不包括这一点
28 因为输入是按照y从小到达，然后是按x从小到达，且x不同
29 因此先出现的肯定比后出现的y小即在后出现的点之下，
30 因此只要把x作为树状数组下标就可以啦。然后用一个out[num]维护
31 个数为num的点的数量
32 */
33 int out[maxn];
34 int main(){
35     int n;
36     while(scanf("%d",&n)!=EOF){
37         int x,y;
38         int i,j;
39         memset(out,0,sizeof(out));
40         memset(C,0,sizeof(C));
41         for(i=1;i<=n;i++){
42             scanf("%d%d",&x,&y);
43             out[sum(x+1)]++;
44             add(x+1,1);
45         }
46         for(i=0;i<n;i++){
47             printf("%d\n",out[i]);
48         }
49     }
50     return 0;
51 }

 


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