题意：
给定两个数组，你可以交换 k 次 两个数组的元素，问最后 a 数组的和最大可以是多少？
题目：
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：
将两个数组sort排序后，分别取前k个，后k个进行比较，如果能使答案变优即可以贪心，交换即可。
AC代码：
#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;
else
swap(a[i],b[n-i+1]);
}
for(i=1; i<=n; i++)
ans+=a[i];
printf("%d\n",ans);
}
return 0;
}