hdoj--2352--Stras（树状数组）

Stars

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                        Total Submission(s): 7381    Accepted Submission(s): 2912

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

51 15 17 13 35 5

 

Sample Output

12110

 

Source

Ural Collegiate Programming Contest 1999

 

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暴力超时了，树状数组吧

//统计每个点左下方的点的个数，输出每一层的个数 #include<cstdio>#include<cstring>#define MAXN 200000int bit[MAXN]; int level[MAXN];//记录每一层 int sum(int x)//求x前边有多少个点 {int ans=0;while(x>0){ans+=bit[x];x-=(x&-x);}return ans;}void add(int x){while(x<=MAXN){bit[x]+=1;x+=(x&-x);}}int main(){int n,x,y;while(scanf("%d",&n)!=EOF){memset(bit,0,sizeof(bit));memset(level,0,sizeof(level));for(int i=0;i<n;i++)//因为输入已经排序的原因，直接构造树状数组{scanf("%d%d",&x,&y);x++;//x是可以等于0的level[sum(x)]++;add(x);//更新bit数组}for(int i=0;i<n;i++)printf("%d\n",level[i]);}return 0;}