B - Two Arrays And Swaps

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本文介绍了一个关于通过交换两个数组中的元素来最大化其中一个数组总和的问题。通过对数组进行特定操作（如记录每个数值出现的频率，并从另一个数组中选择最优值进行交换），可以有效地求解该问题。文章提供了一个具体的实现案例。

You are given two arrays a and b both consisting of n positive (greater than zero) integers. You are also given an integer k.

In one move, you can choose two indices i and j (1≤i,j≤n) and swap ai and bj (i.e. ai becomes bj and vice versa). Note that i and j can be equal or different (in particular, swap a2 with b2 or swap a3 and b9 both are acceptable moves).

Your task is to find the maximum possible sum you can obtain in the array a if you can do no more than (i.e. at most) k such moves (swaps).

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤200) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (1≤n≤30;0≤k≤n) — the number of elements in a and b and the maximum number of moves you can do. The second line of the test case contains n integers a1,a2,…,an (1≤ai≤30), where ai is the i-th element of a. The third line of the test case contains n integers b1,b2,…,bn (1≤bi≤30), where bi is the i-th element of b.

Output
For each test case, print the answer — the maximum possible sum you can obtain in the array a if you can do no more than (i.e. at most) k swaps.

Example
Input

5
2 1
1 2
3 4
5 5
5 5 6 6 5
1 2 5 4 3
5 3
1 2 3 4 5
10 9 10 10 9
4 0
2 2 4 3
2 4 2 3
4 4
1 2 2 1
4 4 5 4

Output

Note
In the first test case of the example, you can swap a1=1 and b2=4, so a=[4,2] and b=[3,1].

In the second test case of the example, you don’t need to swap anything.

In the third test case of the example, you can swap a1=1 and b1=10, a3=3 and b3=10 and a2=2 and b4=10, so a=[10,10,10,4,5] and b=[1,9,3,2,9].

In the fourth test case of the example, you cannot swap anything.

In the fifth test case of the example, you can swap arrays a and b, so a=[4,4,5,4] and b=[1,2,2,1].

做法:
非常暴力的一道题，开两个数组map1，map2，下标代表数值，a[i]代表出现的次数，map1存a，map2存b，然后将b中最大的前k个数存入map1，然后输出map1前n个最大值.

#include<bits\stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int a[100] = { 0 };
        int b[100] = { 0 };
        int n, k;
        int c[40] = { 0 };
        int d[50] = { 0 };
        cin >> n >> k;
        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
            c[a[i]]++;
        }
        for (int i = 0; i < n; i++)
        {
            cin >> b[i];
            d[b[i]]++;
        }
        for (int i = 30; i > 0; i--)
        {
            if (d[i] && k)
            {
                while (d[i] && k)
                {
                    c[i]++;
                    d[i]--;
                    k--;
                }
            }
            if (k <= 0) break;
        }
        int ans = 0;
        for (int i = 30; i > 0; i--)
        {
            if (c[i])
            {
                while (c[i] && n)
                {
                    ans +=i;
                    n--;
                    c[i]--;
                }
            }
            if (n <= 0) break;
        }
        cout << ans << endl;
    }
    return 0;
}