CodeForces 1353 B. Two Arrays And Swaps 貪心算法

1. 題目描述

1.1. Limit

Time Limit: 1 second

Memory Limit: 256 megabytes

1.2. Problem Description

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 $j$ ($1 \le i, j \le n$) and swap $a_i$ and $b_j$ (i.e. $a_i$ becomes $b_j$ and vice versa). Note that $i$ and $j$ can be equal or different (in particular, swap $a_2$ with $b_2$ or swap $a_3$ and $b_9$ 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.

1.3. Input

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

The first line of the test case contains two integers $n$ and $k$ ($1 \le n \le 30; 0 \le k \le n$) — 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 $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 30$), where $a_i$ is the $i$-th element of $a$. The third line of the test case contains $n$ integers $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 30$), where $b_i$ is the $i$-th element of $b$.

1.4. Onput

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.

1.5. Sample 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

1.6. Sample Onput

6
27
39
11
17

1.7. Note

In the first test case of the example, you can swap $a_1 = 1$ and $b_2 = 4$, 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 $a_1 = 1$ and $b_1 = 10$, $a_3 = 3$ and $b_3 = 10$ and $a_2 = 2$ and $b_4 = 10$, 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 $b=[1, 2, 2, 1]$.

1.8. Source

CodeForces 1353 B. Two Arrays And Swaps

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2. 解讀

將 $a$ 序列按照升序排序， 將 $b$ 按照降序排序。

依次遍歷 $a$ 序列中的所有元素，當 $a_i > b_j$時，將 $a_i$ 和 $b_j$ 交換，然後$j = j + 1$。

3. 代碼

#include <algorithm>
#include <iostream>
#include <string.h>
const long long num = 1e2 + 1;
using namespace std;

long long listA[num];
long long listB[num];

// 交換元素
void mySwap(long long& a, long long& b)
{
    long long &c = b;
    b = a;
    a = c;
}
// 降序排列
int cmp(long long a, long long b)
{
    return a > b;
}

int main()
{
    // test case
    long long t;
    long long n, k;
    long long ans = 0;
    long long markA, markB;
    // test case
    scanf("%lld", &t);
    // for each test case
    while (t--) {
        scanf("%lld %lld", &n, &k);
        // 初始化
        memset(listA, 0, sizeof(listA));
        memset(listB, 0, sizeof(listB));
        ans = markA = markB = 0;
        // 輸入
        for (int i = 0; i < n; i++)
            scanf("%lld", &listA[i]);
        for (int i = 0; i < n; i++)
            scanf("%lld", &listB[i]);
        // 排序，A升序B降序
        sort(listA, listA + n);
        sort(listB, listB + n, cmp);
        // 交換
        for (int i = 0; i < k; i++) {
            if (listA[i] < listB[markB]){
                mySwap(listA[i], listB[markB]);
                markB++;
            }
        }

        // 求和
        for (int i = 0; i < n; i++)
            ans += listA[i];

        printf("%lld\n", ans);
    }
}

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