Codeforces Round 945 (Div. 2) C. Cat, Fox and Double Maximum 题解贪心构造_codeforces c. cat, fox and double maximum-优快云博客

Cat, Fox and Double Maximum

题目描述

Fox loves permutations! She came up with the following problem and asked Cat to solve it:

You are given an even positive integer $n$ and a permutation $^\dagger$ $p$ of length $n$ .

The score of another permutation $q$ of length $n$ is the number of local maximums in the array $a$ of length $n$ , where $a_i = p_i + q_i$ for all $i$ ( $\le i \le n$ ). In other words, the score of $q$ is the number of $i$ such that $1 < i < n$ (note the strict inequalities), $a_{i-1} < a_i$ , and $a_i > a_{i+1}$ (once again, note the strict inequalities).

Find the permutation $q$ that achieves the maximum score for given $n$ and $p$ . If there exist multiple such permutations, you can pick any of them.

$^\dagger$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2, 3, 1, 5, 4]$ is a permutation, but $[1, 2, 2]$ is not a permutation ( $2$ appears twice in the array), and $[1, 3, 4]$ is also not a permutation ( $n = 3$ but there is $4$ in the array).

输入描述

The first line of input contains an integer $t$ ( $\leq t \leq 10^4$ ) — the number of test cases in the input you will have to solve.

The first line of each test case contains one even integer $n$ ( $\leq n \leq 10^5$ , $n$ is even) — the length of the permutation $p$ .

The second line of each test case contains the $n$ integers $p_1, p_2, \ldots, p_n$ ( $\leq p_i \leq n$ ). It is guaranteed that $p$ is a permutation of length $n$ .

It is guaranteed that the sum of $n$ across all test cases doesn’t exceed $10^5$ .

输出描述

For each test case, output one line containing any permutation of length $n$ (the array $q$ ), such that $q$ maximizes the score under the given constraints.

样例输入 #1

4
4
1 2 3 4
4
4 3 1 2
6
6 5 1 4 2 3
8
1 2 4 5 7 6 8 3

样例输出 #1

2 4 1 3
3 1 4 2
2 5 1 4 3 6
5 4 8 2 7 1 6 3

原题

CF——传送门
 洛谷——传送门

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vector<int> p(n);
        for (int i = 0; i < n; i++)
        {
            cin >> p[i];
        }
        int n_pos = find(p.begin(), p.end(), n) - p.begin();
        vector<int> q(n);
        // 注意，由于我从下标0开始读入数据，所以在下面的描述中，0,2,4,6……是奇数位，而1,3,5,7……是偶数位
        if (n_pos & 1) // 如果值n在偶数位上，则将p排列翻转，使其变为奇数位上
        {
            reverse(p.begin(), p.end());
        }
        // n在偶数位和n在奇数位的情况最终都转化为n在奇数位的处理
        vector<pair<int, int>> even, odd; // 记录值和位置的数组
        // 先处理偶数位，偶数位需得到1~n/2，小的值交给大的
        for (int i = 1; i < n; i += 2)
        {
            even.push_back({p[i], i});
        }
        sort(even.begin(), even.end(), greater<pair<int, int>>());
        // 按小值交给大值的顺序赋值1~n/2
        for (int i = 0; i < n / 2; i++)
        {
            q[even[i].second] = i + 1;
        }
        // 再处理奇数位，奇数位的第一位得到n/2+1，剩余位得到n/2+2~n，大的值交给小的
        for (int i = 0; i < n; i += 2)
        {
            odd.push_back({p[i], i});
        }
        sort(odd.begin() + 1, odd.end(), greater<pair<int, int>>());
        // 按大值交给小值的顺序赋值n/2+1~n
        for (int i = 0; i < n / 2; i++)
        {
            q[odd[i].second] = i + n / 2 + 1;
        }
        // 若之前翻转过p，则再翻转一次结果q就能得到所需答案
        if (n_pos & 1)
        {
            reverse(q.begin(), q.end());
        }
        for (int i = 0; i < n; i++)
        {
            cout << q[i] << " \n"[i == n - 1]; // 新学到的小寄巧()
        }
    }

    return 0;
}