Code jam Problem A. Minimum Scalar Product

题目描述：

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

解题思路：

本题稍微列举出几个例子，不难发现规律，将第一个向量的最小和第二个的最大相乘，然后次小与次大，以此类推。这题可以用贪心来理解。

但要证明这个思路的正确性却不容易，一开始我打算一步一步地来证，即证明去掉最小和最大乘积，将导出比去掉任意其他乘积更小的值，但可惜失败了。

后来google后发现一个 排序不等式（Rearrangement inequality），wiki上面有对这个定理的详细证明，于是问题就解决了

#include <cstdio>
#include <algorithm>
using namespace std;

int main(){
    //freopen("A-small-practice.in", "r", stdin);
    freopen("A-large-practice.in", "r", stdin);
    //freopen("A-small-practice.out", "w", stdout);
    freopen("A-large-practice.out", "w", stdout);
    int t;
    scanf("%d", &t);
    for(int kase = 1; kase <= t; kase++){
        int cnt;
        scanf("%d", &cnt);
        int v1[cnt+1], v2[cnt+1];
        for(int i = 0; i < cnt; i++)
            scanf("%d", &v1[i]);
        for(int i = 0; i < cnt; i++)
            scanf("%d", &v2[i]);

        sort(v1, v1+cnt);
        sort(v2, v2+cnt);
        long long res = 0;
        //这里要注意转型为long long，否则还是wrong answer
        for(int i = 0; i < cnt; i++)
            res += (long long)v1[i] * (long long)v2[cnt-1-i];
        printf("Case #%d: %lld\n", kase, res);
    }
    return 0;
}