GCJ：2008 Round1AA-Minimum Scalar Product（有序数组倒序乘积和最小）

题目链接：https://code.google.com/codejam/contest/32016/dashboard#s=p0

Minimum Scalar Product

This contest is open for practice. You can try every problem as many times as you like, though we won’t keep track of which problems you solve. Read the Quick-Start Guide to get started.

Problem

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

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 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 ≤ xi, yi ≤ 1000

Large dataset

T = 10

100 ≤ n ≤ 800

-100000 ≤ xi, yi ≤ 100000

Sample

Input

2

3

1 3 -5

-2 4 1

5

1 2 3 4 5

1 0 1 0 1

Output

Case #1: -25

Case #2: 6

解题心得：

第一次做GCJ的题目，这个题目的意思就是给你两个数组，你可以改变数组内任意一个元素的位置，要求a[1]*b[1]+a[2]*b[2]+a[3]*b[3]….a[n]*b[n]的值最小。
刚开始还去讨论判断了一下，其实根本不用讨论，就是有序数组倒序乘积和最小，也就是sum(a[i]*b[n-i+1])的值。下面进行了部分证明，其实拓展之后也是适用的，懒得写了。

代码：

#include <algorithm>

#include <stdio.h>

#include <sudo_plugin.h>

using namespace std;

typedef long long ll;

const int  maxn = 1000;

ll num1[maxn],num2[maxn];

int main() {

    FILE *fp,*fout;

    fp = fopen("A-large-practice.in","r");

    fout = fopen("A.out","w");

    int t,T = 1;

    fscanf(fp,"%d",&t);

    while(t--) {

        int n;

        fscanf(fp,"%d",&n);

        for (int i = 0; i < n; i++)

            fscanf(fp,"%lld",&num1[i]);

        for (int i = 0; i < n; i++)

            fscanf(fp,"%lld",&num2[i]);

        sort(num1, num1 + n);

        sort(num2, num2 + n);

        int r = n - 1;

        ll sum = 0;

        for (int i = 0; i < n; i++) {

            sum += num1[i] * num2[r--];

        }

        fprintf(fout,"Case #%d: %lld\n",T++ , sum);

    }

    fclose(fout);

    fclose(fp);

    return 0;

}

巴特西