Google Code Jam Notes - Minimum Scalar Product - Java

最新推荐文章于 2021-02-27 06:52:49 发布

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最新推荐文章于 2021-02-27 06:52:49 发布

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分类专栏： code jam 文章标签： google code jam java minimum scalar produ

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本文探讨了如何通过重新排列两个向量的元素来获得最小化的标量乘积。介绍了问题背景、输入输出格式及样例，并提供了一种有效的算法实现，通过排序向量元素并按逆序配对的方法来求得最小标量乘积。

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Problem:

Retrieved from: http://code.google.com/codejam/contest/32016/dashboard#s=p0

You are given two vectors v₁=(x₁,x₂,...,x_n) and v₂=(y₁,y₂,...,y_n). The scalar product of these vectors is a single number, calculated as x₁y₁+x₂y₂+...+x_ny_n.

Suppose you are allowed to permute the coordinates of each vector as you wish. Choose two permutations such that the scalar product of your two new vectors is the smallest possible, and output that minimum scalar product.

Input

The first line of the input file contains integer number T - the number of test cases. For each test case, the first line contains integer number n . The next two lines contain n integers each, giving the coordinates of v ₁ and v ₂ respectively.

Output

For each test case, output a line

Case #X: Y

where X is the test case number, starting from 1, and Y is the minimum scalar product of all permutations of the two given vectors.

Limits

Small dataset

T = 1000
1 ≤ n ≤ 8
-1000 ≤ x_i, y_i ≤ 1000

Large dataset

T = 10
100 ≤ n ≤ 800
-100000 ≤ x_i, y_i ≤ 100000

Sample

Input	Output
`2 3 1 3 -5 -2 4 1 5 1 2 3 4 5 1 0 1 0 1`	`Case #1: -25 Case #2: 6`

Analysis:

This is actually a mathematical problem, the fact is if we have two arrays in natural order:

(x1, x2, x3, x4) and (y1, y2, y3, y4)

The minimum scalar product is: x1* y4 + x2*y3 + x3*y2 + x4*y1

The maximum scalar product is: x1*y1 + x2*y2 + x3* y3 + x4*y4

My solution: (Your opinion is highly appreciated)

package codeJam.google.com;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.Arrays;

/**
 * @author Zhenyi 2013 Dec 21, 2013 14:45:21 PM
 */
public class MinimumScalarProduct {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new FileReader(
				"C:/Users/Zhenyi/Downloads/A-small-practice.in"));
		FileWriter out = new FileWriter(
				"C:/Users/Zhenyi/Downloads/A-small-practice.out");
		// BufferedReader in = new BufferedReader(new
		// FileReader("C:/Users/Zhenyi/Downloads/A-large-practice.in"));
		// FileWriter out = new
		// FileWriter("C:/Users/Zhenyi/Downloads/A-large-practice.out");

		int N = new Integer(in.readLine());

		for (int cases = 1; cases <= N; cases++) {
			int num = new Integer(in.readLine());
			String[] sta = new String(in.readLine()).split("\\s");
			String[] stb = new String(in.readLine()).split("\\s");
			long a[] = new long[num];
			for (int i = 0; i < num; i++) {
				a[i] = new Long(sta[i]);
			}

			long b[] = new long[num];
			for (int i = 0; i < num; i++) {
				b[i] = new Long(stb[i]);
			}

			long result = 0;

			Arrays.sort(a);
			Arrays.sort(b);

			for (int i = 0; i < a.length; i++) {
				result = result + a[i] * b[b.length - i - 1];
			}

			out.write("Case #" + cases + ": " + result + "\n");

		}
		in.close();
		out.flush();
		out.close();
	}
}