poj-2352【Stars】

Stars

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42415 Accepted: 18470

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

题意：每一个坐标代表着每一个星星的位置，位于这颗星星的左下方星星的个数代表着
这颗星星的等级，(除去它自己);输出等级是0~n-1的星星的个数 
刚学的树状数组，网上说是基础题,(你们在坑我吗？),想了好长时间才明白它的本质，其实主要
是题目的输入按照y递增，y相同时按照x的递增输入，我们就不理理会y了，对于tree数组的下标就是按照
横坐标的顺序排的，对于你输入的x的值，那右端的肯定等级需要+1；无论y值有多大，都行，因为，是按照
递增输入的，仔细想想，只能说到这了

#include<cstdio>#include<cstring>#define maxn 35000int level[20000];int tree[maxn];int lowbit(int x){return x&-x;}void add(int x){while(x<maxn){tree[x]++;x+=lowbit(x);}}int get_sum(int x){int sum=0;while(x){sum+=tree[x];x-=lowbit(x);}return sum;}int main(){int n;scanf("%d",&n);int a=n;while(n--){int x,y;scanf("%d%d",&x,&y);++x;//防止出现x为0的情况level[get_sum(x)]++;add(x); }for(int i=0;i<a;++i)printf("%d\n",level[i]);return 0;}