poj 2352 Stars

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26076		Accepted: 11394

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n个星星的坐标，每个星星都有一个等级，等级为x坐标和y坐标都不大于该星星的星星数目。求每个等级的星星数目。

思路：树状数组求区间和。y坐标是递增给出的，每次加入一个星星(x,y)，区间[1,x]的总和（注意此时不包含要加入的点）即为该星星的级数，然后更新树状数组。

 

AC代码：

#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <iostream>
#define max2(a,b) ((a) > (b) ? (a) : (b))
#define min2(a,b) ((a) < (b) ? (a) : (b))

using namespace std;
const int N=32005;
int n;
int c[N],cnt[N];
int lowbit(int x)
{
    return x&(-x);
}
void modify(int x,int d)
{
    for(int i=x;i<=32005;i+=lowbit(i))
    c[i]+=d;
}
int sum(int x)
{
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i))
        ret+=c[i];
    return ret;
}
int main()
{
    int x,y;
    memset(c,0,sizeof(c));
    memset(cnt,0,sizeof(cnt));
   scanf("%d",&n);
   for(int i=0;i<n;i++)
   {
       scanf("%d%d",&x,&y);
       int ans=sum(x+1);
       cnt[ans]++;
       modify(x+1,1);
   }
   for(int i=0;i<n;i++)
   printf("%d\n",cnt[i]);
   return 0;
}