【ICPC】The 2021 CCPC Guilin Onsite (XXII Open Cup, Grand Prix of EDG) D

Assumption is All You Need

#模拟 #构造

题目描述

JB holds the belief that assumption is all you need to solve a problem. In order to prove that, JB has given you two permutations of numbers from $1$ to $n$: $A$ and $B$, and JB wants you to output a sequence of element swapping operation $(x_i,y_i)$ on $A$, so that:

1. every pair of swapped element forms an inversion pair ($i.e.$ $x_i\le y_i\le$and $A_{x_i}\ge A_{y_i}$ when the $i$-th operation is being performed)
2. $A$ will become $B$ at the end of the swapping sequence.

or determine it is impossible. Help prove JB’s belief by solving this problem!

输入格式

There are multiple test cases. The first line of the input contains one integer $T$ indicating the number of test cases. For each test case:

The first line contains one integer $n$ ($1\le n\le 2021$), indicating the number elements in $A$ and $B$.

The second line contains $n$ distinct integers $A_1,A_2,\dots ,A_n$ ($1\le A_i\le n$), indicating the array $A$.

The third line contains $n$ distinct integers $B_1,B_2,\dots ,B_n$ ($1\le B_i\le n$), indicating the array $B$.

It is guaranteed that the sum of $n$ in all test cases will not exceed $2021$.

输出格式

For each test case, if there doesn’t exist a sequence, output the one line containing one integer “-1”.

Otherwise, in the first line output one integer $k$ ($0\le k\le \frac{n(n-1)}{2}$), indicating the length of the swapping sequence. Then, output $k$ line each containing two integers $x_i$ and $y_i$ ($1\le x_i\le y_i\le n$), indicating the $i$-th operation $\text{swap}(A_{x_i},A_{y_i})$.

样例 #1

样例输入 #1

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

样例输出 #1

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

解法

观察到$n\le 10^3$，并且最多交换$\frac{n(n-1)}{2}$ 次，容易想到$n^2$模拟。

而每次交换只能交换逆序对，把值较小的交换到前面来，因此我们要尽可能的让大的数放到前面来，这样才更有机会被交换。

因此，最优的交换策略之一，就是$i$之后的$j(j>i)$满足$a_j<a_i\cap a_j\le b_i$的$a_j$换到前面来，如果交换之后，仍然出现$a_i<a_j$的情况，则无解。

代码

void solve() {int n;std::cin >> n;std::vector<int>a(n + 1), b(n + 1);for (int i = 1; i <= n; ++i) {std::cin >> a[i];}for (int i = 1; i <= n; ++i) {std::cin >> b[i];}std::vector<pii>res;for (int i = 1; i <= n; ++i) {if (a[i] == b[i]) continue;if (a[i] < b[i]) {std::cout << -1 << "\n";return;}int j = i + 1;while (j <= n && a[i] != a[j]) {if (a[j] < a[i] && a[j] <= b[i]) {std::swap(a[i], a[j]);res.push_back({ i,j });}j++;}}std::cout << res.size() << "\n";for (auto& [x, y] : res) {std::cout << x << " " << y << "\n";}}signed main() {std::ios::sync_with_stdio(0);std::cin.tie(0);std::cout.tie(0);int t = 1;std::cin >> t;while (t--) {solve();}return 0;
}