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2013
11-11

POJ 2352 Stars [解题报告] Java

Stars

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

5
1 1
5 1
7 1
3 3
5 5

样例输出:

1
2
1
1
0

温馨提示:

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

解题代码:

解法一(用树状数组)
import java.io.*;
public class Main
{
	static int[] star=new int[32002];
	static int[] lev=new int[15000];
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		InputStreamReader is=new InputStreamReader(System.in);
		BufferedReader in=new BufferedReader(is);
		int a=Integer.parseInt(in.readLine());
		int n=a;
		while((a--)!=0)
		{
			String[] ss=in.readLine().split(" ");
			int b=Integer.parseInt(ss[0]);
			lev[sum(b+1)]++;
			update(b+1);
		}
		for(int i=0;i< n;i++)
			System.out.println(lev[i]);
	}

	static int lowbit(int n)
	{
		return n&(-n);
	}

	static int sum(int n)
	{
		int r=0;
		while(n!=0)
		{
			r+=star[n];
			n-=lowbit(n);
		}
		return r;
	}
	
	static void update(int n)
	{
		while(n< 32002)
		{
			star[n]++;
			n+=lowbit(n);
		}
	}
}

解法二:
//* @author: [email protected]
import java.io.*;
public class Main
{
 static int[] tree,ans;
 public static void main(String[] args) throws IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  int a=Integer.parseInt(in.readLine());
  int n=a;
  tree=new int[32000*3];
  ans=new int[a];
  while((a--)!=0)
  {
   String[] ss=in.readLine().split(" ");
   int x=Integer.parseInt(ss[0]);
   int u=search(1,x,0,32000);
   ans[u]++;
  }
   for(int i=0;i< n;i++)
	System.out.println(ans[i]);
 }

 static int search(int i,int x,int l,int r)
 {
  tree[i]++;
  if(x==r) return tree[i]-1;
  int mid=(l+r)/2;
  if(x<=mid) return search(i*2,x,l,mid);
  else return tree[i*2]+search(i*2+1,x,mid+1,r);
 }
}

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮