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2013
11-11

POJ 2460 Brownie Points I [解题报告] Java

Brownie Points I

问题描述 :

Stan and Ollie play the game of Odd Brownie Points. Some brownie points are located in the plane, at integer coordinates. Stan plays first and places a vertical line in the plane. The line must go through a brownie point and may cross many (with the same x-coordinate). Then Ollie places a horizontal line that must cross a brownie point already crossed by the vertical line.

Those lines divide the plane into four quadrants. The quadrant containing points with arbitrarily large positive coordinates is the top-right quadrant.

The players score according to the number of brownie points in the quadrants. If a brownie point is crossed by a line, it doesn’t count. Stan gets a point for each (uncrossed) brownie point in the top-right and bottom-left quadrants. Ollie gets a point for each (uncrossed) brownie point in the top-left and bottom-right quadrants.

Your task is to compute the scores of Stan and Ollie given the point through which they draw their lines.

输入:

Input contains a number of test cases. The data of each test case appear on a sequence of input lines. The first line of each test case contains a positive odd integer 1 < n < 200000 which is the number of brownie points. Each of the following n lines contains two integers, the horizontal (x) and vertical (y) coordinates of a brownie point. No two brownie points occupy the same place. The input ends with a line containing 0 (instead of the n of a test).

输出:

For each test case of input, print a line with two numbers separated by a single space. The first number is Stan’s score, the second is the score of Ollie when their lines cross the point whose coordinates are given at the center of the input sequence of points for this case.

样例输入:

11
3 2
3 3
3 4
3 6
2 -2
1 -3
0 0
-3 -3
-3 -2
-3 -4
3 -7
0

样例输出:

6 3

解题代码:

//* @author: ccQ.SuperSupper
import java.io.*;
class point{
	int x,y;
}
public class Main {
 static int n;
 static final int N = 200000+100;
 static point position[] = new point[N];
	
 public static void main(String[]args) throws Exception{
		
  int i;
  start();
		
  StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
  while(true){
	n = Get_Num(cin);
	if(n<=0) break;
	for(i=0;i< n;++i){
		position[i].x = Get_Num(cin);
		position[i].y = Get_Num(cin);
	}
	solve();
  }
 }

 static void start(){
  for(int i=0;i< N;++i) position[i] = new point();
  }

  static void solve(){
  int i,x,y,ansa,ansb;
  ansa = ansb = 0;
  x = position[n/2].x;
  y = position[n/2].y;
  for(i=0;i< n;++i){
	if((position[i].x>x && position[i].y>y)||(position[i].x< x && position[i].y< y))
		++ansa;
	if((position[i].x>x && position[i].y< y)||(position[i].x< x && position[i].y>y))
		++ansb;
  }
  System.out.println(ansa+" "+ansb);
 }

 static int Get_Num(StreamTokenizer cin) throws Exception{
	cin.nextToken();
	return (int)cin.nval;
 }
}

  1. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?