poj 2352

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25160		Accepted: 10983

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.

题意：给出一些点的坐标，在该点下方和左方的点个数为0,1，……n-1的个数。

思路：树状数组存储，按照x计算个数。因为给出的坐标就是按照顺序给出的 。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define N 32005
int c[N];
int du[N];

int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int val)
{
    while(x<=N)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}
int getsum(int x)
{
    int result=0;
    while(x>0)
    {
        result+=c[x];
        x-=lowbit(x);
    }
    return result;
}
int main()
{
    int n,x,y;
    memset(c,0,sizeof(c));
    memset(du,0,sizeof(du));
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&x,&y);//由于y已经是从小到大排好序了，所以只要由x计数即可。
        int d=getsum(x+1);
        du[d]++;
        add(x+1,1);
    }
    for(int i=0;i<n;i++)
    printf("%d\n",du[i]);
    return 0;
}