Stars

1028. Stars

Time limit: 0.25 second
Memory limit: 64 MB

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 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 ofY 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

inputoutput

51 15 17 13 35 5

12110

/**线段树**/#include <iostream>using namespace std;#define maxn    16005#define lson    l,m,root<<1#define rson    m+1,r,root<<1^1int level[maxn]={0};int sum[maxn<<2+1]={0};int X,Y;int n;void PushUp(int root){    sum[root]=sum[root<<1]+sum[root<<1|1];}void Update(int l,int r,int root)/**同上一题**/{    if(l==X&&r==X)    {        sum[root]++;        return ;    }    int m=(l+r)>>1;    if(X<=m)        Update(lson);    else        Update(rson);    PushUp(root);}int query(int L,int R,int l,int r,int root){    if(L>=l&&r<=R)        return sum[root];    int m=(l+r)>>1;    int ret=0;    /**/    if(R<=m)/**这里与上一题“炮兵阵地”有所不同**/        ret=query(L,R,lson);/**如果L,R在左孩子区间里,就在左子树的区间查找**/    else if(L>m)        ret=query(L,R,rson);/**如果L,R在右孩子区间里,就在右子树的区间查找**/    else        ret=query(L,m,lson)+query(m+1,R,rson);/**如果左右孩子区间都各有一部分,只需要将两部分相加并返回**/    return ret;}void read_deal()/**由于题目所给的坐标是已经排好顺序的了，因此只需要读入横坐标就行；以后这类的也一定要注意顺序**/{    cin>>n;    for(int i=0;i<n;i++)    {        cin>>X>>Y;/** 注意查询区间,是从0~x,包括x **/        int temp=query(0,X,0,16000*2,1);/**因为题目所说的等级计算方法是不包括自己的，所以再输入每个点之前查询**/        level[temp]++;        Update(0,16000*2,1);    }}void print(){    for(int i=0;i<n;i++)        cout<<level[i]<<endl;}int main(){    read_deal();    print();    return 0;}