AtCoder Regular Contest 153

B - Grid Rotations

https://atcoder.jp/contests/arc153/tasks/arc153_b

Time Limit: $2$ sec / Memory Limit: $1024$ MB

Score : $500$ points

Problem Statement

We have a grid with $H$ rows from top to bottom and $W$ columns from left and right. Initially, the square at the $i$-th row from the top and $j$-th column from the left has a lowercase English letter $A_{i,j}$.

Let us perform $Q$ operations on this grid. In the $i$-th operation, we are given integers $a_i$ and $b_i$ such that $1\leq a_i \leq H-1$ and $1\leq b_i\leq W-1$, and do the following.

  • Let $R_1$, $R_2$, $R_3$, and $R_4$ be rectangular regions within the grid defined as follows:

    • $R_1$ is the intersection of the top $a_i$ rows and leftmost $b_i$ columns;
    • $R_2$ is the intersection of the top a_i rows and rightmost $W-b_i$ columns;
    • $R_3$ is the intersection of the bottom H-a_i rows and leftmost $b_i$ columns;
    • $R_4$ is the intersection of the bottom H-a_i rows and rightmost $W-b_i$ columns.
  • Rotate $180$ degrees each of $R_1$, $R_2$, $R_3$, and $R_4$.
    Here, a $180$-degree rotation of a rectangular region $R$ within the grid moves the character on the square at the $i$-th from the top and $j$-th column from the left in $R$ to the square at the $i$-th from the bottom and $j$-th column from the right in $R$. See also the figures for the samples.

Print the grid after all $Q$ operations.

Constraints

  • $2\leq H, W$, and $HW \leq 5\times 10^5$.
  • $A_{i,j}$ is a lowercase English letter.
  • $1\leq Q\leq 2\times 10^5$
  • $1\leq a_i\leq H - 1$
  • $1\leq b_i\leq W - 1$

Input

The input is given from Standard Input in the following format:

$H$ $W$
$A_{1,1}\cdots A_{1, W}$
$\vdots$
$A_{H,1}\cdots A_{H, W}$
$Q$
$a_1$ $b_1$
$\vdots$
$a_Q$ $b_Q$

Output

Print the grid after the operations in the following format, where $B_{i,j}$ is the character on the square $(i,j)$ on the final grid.

$B_{1,1}\cdots B_{1, W}$
$\vdots$
$B_{H,1}\cdots B_{H, W}$

Sample Input 1

4 5
abcde
fghij
klmno
pqrst
1
3 3

Sample Output 1

mlkon
hgfji
cbaed
rqpts

The grid will change as follows.

Sample Input 2

3 7
atcoder
regular
contest
2
1 1
2 5

Sample Output 2

testcon
oderatc
ularreg

The grid will change as follows.

Sample Input 3

2 2
ac
wa
3
1 1
1 1
1 1

Sample Output 3

ac
wa

The grid will change as follows.

题解

同一行/列元素操作后仍在同一行/列

通过观察可以发现，题目所述操作是对整行、整列进行的。因此同一行元素操作后仍在同一行、同一列元素操作后仍在同一列，只是顺序变化了。

由此，我们可以想到，无需维护每次操作整个矩阵的值，而只用维护行列前后对应关系即可，即第 $x$ 行/列操作后变成了第 $y$ 行/列。最后输出时根据行列对应关系生成新矩阵即可。

如果使用暴力法，时间复杂度为：$O(Q\cdot(H+W))$. 根据题目要求，最坏的情况是 $H=W\approx700$，复杂度 $10^8$ 级别，会超时。由此可见，必须对行列对应关系的维护方式进行优化。

行列对应关系的特点

我们通过一个例子来解释，我们先只考虑行，因为这个问题行列是独立的：

$$ \begin{align} 初始情况:\;&12345678\\ 对第4位操作:\;&43218765\\ 对第2位操作:\;&34567812\\ 对第7位操作:\;&18765432\\ 对第3位操作:\;&78123456 \end{align} $$

此时获取还看不出规律，我们加上一些辅助：

$$ \begin{align} 初始情况:\;&\overrightarrow{12345678}|\\ 对第4位操作:\;&\overleftarrow{4321}|\overleftarrow{8765}\\ 对第2位操作:\;&\overrightarrow{345678}|\overrightarrow{12}\\ 对第7位操作:\;&\overleftarrow{1}|\overleftarrow{8765432}\\ 对第3位操作:\;&\overrightarrow{78}|\overrightarrow{123456} \end{align} $$

这样就很明显了，每次操作其实序列都是顺序的，只不过起始点发生了偏移。并且奇数次操作时为逆序，偶数次操作时为顺序。那么我们就不需要维护整个行列对应关系了，而只需要维护起始点 $1$ 就行了。最后输出时，通过起始点的位置，结合操作的次数，生成出行列对应关系，再用行列对应关系生成出新矩阵就行了。

时间复杂度为：$O(Q+HW)$，可以满足题目要求。

代码

#include <bits/stdc++.h>
#define endl '\n'

using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int H, W;
    cin >> H >> W;
    vector<string> A(H);
    for (int i = 0; i < H; i++)
        cin >> A[i];
    int Q;
    cin >> Q;
    int start_row = 0, start_col = 0;
    for (int i = 0; i < Q; i++)
    {
        int a, b;
        cin >> a >> b;
        a--, b--;
        start_row = (start_row <= a) ? (a - start_row) : (H - start_row + a);
        start_col = (start_col <= b) ? (b - start_col) : (W - start_col + b);
    }
    vector<int> row(H), col(W);
    if (Q & 1)
    {
        for (int i = 0; i < H; i++)
            row[(start_row - i + H) % H] = i;
        for (int i = 0; i < W; i++)
            col[(start_col - i + W) % W] = i;
    }
    else
    {
        for (int i = 0; i < H; i++)
            row[(start_row + i) % H] = i;
        for (int i = 0; i < W; i++)
            col[(start_col + i) % W] = i;
    }
    for (int i = 0; i < H; i++)
    {
        for (int j = 0; j < W; j++)
            cout << A[row[i]][col[j]];
        cout << endl;
    }
    return 0;
}