HDU1541(树状数组)

　　

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8371    Accepted Submission(s): 3337

Problem 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

题目大意： 开始输入一个n（表示有n颗星星）。接下来n行，每行输入一个x，y，代表星星的坐标，注意y是递增输入。求各阶段星星的等级（阶段分为0~n-1）有多少，星星等级数计算方法为本星星左下角的有的星星个数。

思路：因为y递增，只需判断有多少点的横坐标小于当前的ｘ，即可得知当前星星的等级。可用树状数组维护ｘ。

//2016.8.10
#include<iostream>
#include<cstdio>
#include<cstring>
#define N 32005

using namespace std;

int n, x, y, arr[N], level[N];//arr[i]表示坐标为ｘ的点出现的次数，level[i]表示第i级星星的数目

int lowbit(int x){
    return x&(-x);
}

int add(int pos, int num)
{
    for(int i = pos; i <= N; i+=lowbit(i))
      arr[i]+=num;
}

int query(int l)
{
    int sum = 0;
    for(int i = l; i > 0; i-=lowbit(i))
      sum += arr[i];
    return sum;
}

int main()
{
    while(cin>>n)
    {
        memset(arr, 0, sizeof(arr));
        memset(level, 0, sizeof(level));
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &x, &y);
            level[query(x+1)]++;//因为x可以等于0，为防止出现死循环，将横坐标整体向右平移一格
            add(x+1, 1);
        }

        for(int i = 0; i < n; i++)
          cout<<level[i]<<endl;
    }

    return 0;
}