E. Working routine
time limit per test
2.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasiliy finally got to work, where there is a huge amount of tasks waiting for him. Vasiliy is given a matrix consisting of n rows and mcolumns and q tasks. Each task is to swap two submatrices of the given matrix.
For each task Vasiliy knows six integers ai, bi, ci, di, hi, wi, where ai is the index of the row where the top-left corner of the first rectangle is located, bi is the index of its column, ci is the index of the row of the top-left corner of the second rectangle, di is the index of its column, hi is the height of the rectangle and wi is its width.
It's guaranteed that two rectangles in one query do not overlap and do not touch, that is, no cell belongs to both rectangles, and no two cells belonging to different rectangles share a side. However, rectangles are allowed to share an angle.
Vasiliy wants to know how the matrix will look like after all tasks are performed.
Input
The first line of the input contains three integers n, m and q (2 ≤ n, m ≤ 1000, 1 ≤ q ≤ 10 000) — the number of rows and columns in matrix, and the number of tasks Vasiliy has to perform.
Then follow n lines containing m integers vi, j (1 ≤ vi, j ≤ 109) each — initial values of the cells of the matrix.
Each of the following q lines contains six integers ai, bi, ci, di, hi, wi (1 ≤ ai, ci, hi ≤ n, 1 ≤ bi, di, wi ≤ m).
Output
Print n lines containing m integers each — the resulting matrix.
Examples
input
Copy
4 4 2 1 1 2 2 1 1 2 2 3 3 4 4 3 3 4 4 1 1 3 3 2 2 3 1 1 3 2 2
output
Copy
4 4 3 3 4 4 3 3 2 2 1 1 2 2 1 1
input
Copy
4 2 1 1 1 1 1 2 2 2 2 1 1 4 1 1 2
output
Copy
2 2 1 1 2 2 1 1
题意：给一个矩阵，有q次操作，每次交换两个不相交的子矩阵
可以发现，每次交换时子矩阵内部位置关系是不变的，所以只要交换两个子矩阵的外部位置就行了，每个点开一个四个方向的链表，存每个方向的点，每次交换只要交换链表即可，具体看代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxm = 1005;
struct node
{
int x, y;
node() {};
node(int x, int y) :x(x), y(y) {};
}U[maxm][maxm], D[maxm][maxm], L[maxm][maxm], R[maxm][maxm];
int f[maxm][maxm];
int main()
{
int n, i, j, k, sum, m, Q;
int a, b, c, d, h, w;
node u, v, p, q;
scanf("%d%d%d", &n, &m, &Q);
for (i = 1;i <= n;i++)
for (j = 1;j <= m;j++)
scanf("%d", &f[i][j]);
for (i = 0;i <= n + 1;i++)
{
for (j = 0;j <= m + 1;j++)
{
U[i][j] = node(i - 1, j), D[i][j] = node(i + 1, j);
L[i][j] = node(i, j - 1), R[i][j] = node(i, j + 1);
}
}
while (Q--)
{
scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &h, &w);
u.x = a, u.y = 0, v.x = c, v.y = 0;
for (i = 1;i <= b;i++) u = R[u.x][u.y];
for (i = 1;i <= d;i++) v = R[v.x][v.y];
for (i = 1;i <= w;i++)
{
p = U[u.x][u.y], q = U[v.x][v.y];
U[u.x][u.y] = q, U[v.x][v.y] = p;
D[p.x][p.y] = v, D[q.x][q.y] = u;
if (i < w) u = R[u.x][u.y], v = R[v.x][v.y];
}
for (i = 1;i <= h;i++)
{
p = R[u.x][u.y], q = R[v.x][v.y];
R[u.x][u.y] = q, R[v.x][v.y] = p;
L[p.x][p.y] = v, L[q.x][q.y] = u;
if (i < h) u = D[u.x][u.y], v = D[v.x][v.y];
}
for (i = 1;i <= w;i++)
{
p = D[u.x][u.y], q = D[v.x][v.y];
D[u.x][u.y] = q, D[v.x][v.y] = p;
U[p.x][p.y] = v, U[q.x][q.y] = u;
if (i < w) u = L[u.x][u.y], v = L[v.x][v.y];
}
for (i = 1;i <= h;i++)
{
p = L[u.x][u.y], q = L[v.x][v.y];
L[u.x][u.y] = q, L[v.x][v.y] = p;
R[p.x][p.y] = v, R[q.x][q.y] = u;
if (i < h) u = U[u.x][u.y], v = U[v.x][v.y];
}
}
for (i = 1;i <= n;i++)
{
p.x = i, p.y = 0;
for (j = 1;j <= m;j++)
{
p = R[p.x][p.y];
printf("%d%c", f[p.x][p.y], j == m ? '\n' : ' ');
}
}
return 0;
}