題意：

給定兩個數組，你可以交換 k 次 兩個數組的元素，問最后 a 數組的和最大可以是多少？

題目：

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

6
27
39
11
17

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].

分析：

將兩個數組sort排序后，分別取前k個，后k個進行比較，如果能使答案變優即可以貪心，交換即可。

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int a[35],b[35]; int t,n,k; int main() {scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);int i,ans=0;for(i=1; i<=n; i++)scanf("%d",&a[i]);for(i=1; i<=n; i++)scanf("%d",&b[i]);sort(a+1,a+n+1);sort(b+1,b+n+1);for(i=1; i<=k; i++){if(a[i]>b[n-i+1])break;elseswap(a[i],b[n-i+1]);}for(i=1; i<=n; i++)ans+=a[i];printf("%d\n",ans);}return 0; }

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