(code jam)Problem A. Minimum Scalar Product_考虑两个向量 v1=(x1,x2,...,xn)和v2=(y1,y2,...,yn),向量的内积定义-优快云博客

本文链接：https://blog.youkuaiyun.com/modiz/article/details/21185639

本文介绍了一道算法题目，要求通过适当排列两个向量的元素来最小化它们的点积。提供了完整的C++代码实现，并采用了一种简单有效的策略：将一个向量升序排序，另一个向量降序排序，以此确保最小的点积结果。

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Problem

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 nintegers each, giving the coordinates of v₁ and v₂ 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`

题意：就是给出两个向量的坐标,比如x(x1,x2,x3...) y(y1,y2,y3)任意交换两向量之间的值，使x1*y1+x2*y2+x3*y3...的和最小。

题解就是两个排序，一个从小到大，一个从大到小相乘即是最小值。注意的是大数的时候__in64会超范围。所以用double。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
bool cmp(double a,double b)
{
	return a>b;
}
bool cmp1(double a,double b)
{
	return b>a;
}
double x[1000],y[1000];
int main()
{
	int T,n,i,t=0;
	freopen("A-large-practice.in","r",stdin);
	freopen("output.out","w",stdout);
	scanf("%d",&T);
	while (T--)
	{
		printf("Case #%d: ",++t);
		scanf("%d",&n);
		for (i=0;i<n;i++)
			scanf("%lf",&x[i]);
		for (i=0;i<n;i++)
			scanf("%lf",&y[i]);
		sort(x,x+n,cmp);
		sort(y,y+n,cmp1);
		double s=0;
		for (i=0;i<n;i++)
			s+=x[i]*y[i];
		printf("%.0lf\n",s);
	}
	return 0;
}