poj 2352 Stars

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
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：给你一张星星地图，每个星星的等级是在它左下方星星的个数，求每个等级对应的星星的个数。
做法： 线段树，每次输入的时候查询前面有几个星星的横坐标小于他（因为星星的顺序是y升序的 所以前面的一定是小于等于它），然后更新。

#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>#define LL long long#define maxn 32100using namespace std;int level[4*maxn];int tree[4*maxn];void update(int p,int l,int r,int x){    if(l == r)    {        tree[p]++;        return ;    }    int mid = (l + r) >> 1;    if(x <= mid) update(p*2,l,mid,x);    else         update(p*2+1,mid+1,r,x);    tree[p] = tree[p*2] + tree[p*2+1];}int query(int p,int l,int r,int x,int y){    if(l >= x && r <=y)    {        return tree[p];    }    int mid = (l + r) >> 1;    if(y <= mid)    {        return query(p*2,l,mid,x,y);    }    else if(x > mid)    {        return query(p*2+1,mid+1,r,x,y);    }    else        return query(p*2,l,mid,x,mid)+query(p*2+1,mid+1,r,mid+1,y);}int main(){    int n;    while(~scanf("%d",&n))    {        memset(tree,0,sizeof(tree));        memset(level,0,sizeof(level));        for(int i = 1; i <= n; i++)        {            int x,y;            scanf("%d%d",&x,&y);            x++;            int temp = query(1,1,maxn,1,x);            level[temp]++;            update(1,1,maxn,x);        }        for(int i = 0; i <= n-1; i++)        {            printf("%d\n",level[i]);        }    }    return 0;}