Codeforces Round #367 (Div. 2) E. Working routine （十字链表）

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

4 4 21 1 2 21 1 2 23 3 4 43 3 4 41 1 3 3 2 23 1 1 3 2 2

output

4 4 3 34 4 3 32 2 1 12 2 1 1

input

4 2 11 11 12 22 21 1 4 1 1 2

output

2 21 12 21 1

题意：给你一个n*m的矩阵，然后询问q次，每次询问中的 lx，ly，rx，ry，h，w，分别表示两个子矩阵的左上角坐标和其长宽，要你每次询问要你交换这两个子矩阵。经过q次询问之后，要你输出最后的矩阵。

题解：（看了CF题解才会写啊。。。）

当你交换子矩阵时，它子矩阵的结构实际上是不变的。如图。（橙色是左邻居，红色是右邻居，蓝色是上邻居，绿色是下邻居）

假如我们想要交换2x2的子矩阵（1,2,5,6）和子矩阵（11,12,15,16）。交换后，我们会发现它们的整体并没有改变，因为这两个矩阵仍然有相同的元素。但他们的外边界变了，因为交换时邻居变了，所以，我们可以通过修改边界来达到目的。

先弄个十字链表，然后不断修改邻居（边界）。

代码：

#pragma comment(linker, "/STACK:102400000,102400000")//#include<bits/stdc++.h>#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<map>#include<cmath>#include<queue>#include<set>#include <utility>#include<stack>using namespace std;typedef long long ll;typedef unsigned long long ull;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)  #define rep(i,j,k) for (int i = j; i <= k; i++)  #define per(i,j,k) for (int i = j; i >= k; i--)  #define lson x << 1, l, mid  #define rson x << 1 | 1, mid + 1, r  const int lowbit(int x) { return x&-x; }  const double eps = 1e-8;  const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9 + 7;  const ll mod = (1LL<<32);const int N = 1010;const int M=2000010; template <class T1, class T2>inline void getmax(T1 &a, T2 b) {if(b>a)a = b;}  template <class T1, class T2>inline void getmin(T1 &a, T2 b) {if(b<a)a = b;}int read(){int v = 0, f = 1;char c =getchar();while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();return v*f;}int a[N][N];int r[M],d[M];int main() {    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int n,m;    int q,h,w,lx,ly,rx,ry;    n=read();m=read();q=read();    for(int i=1; i<=n; i++)    for(int j=1; j<=m; j++)    {    a[i][j]=read();}for(int i=0;i<=n;i++){for(int j=0;j<=m;j++){r[i*(m+1)+j] = i*(m+1)+(j+1);d[i*(m+1)+j] = (i+1)*(m+1)+j;}}/*cout<<"-------------------------------------"<<endl;for(int i=0;i<=n;i++){for(int j=0;j<=m;j++){printf("r[%d]=%d\nd[%d]=%d\n",i,r[i*(m+1)+j],i,d[i*(m+1)+j]);cout<<"--------"<<endl;}}cout<<"-------------------------------------"<<endl;*/while(q--){scanf("%d%d%d%d%d%d",&lx,&ly,&rx,&ry,&h,&w);int up = 0;int down = 0;for(int i=1;i<lx;i++) up=d[up];for(int i=1;i<ly;i++) up=r[up];for(int i=1;i<rx;i++) down=d[down];for(int i=1;i<ry;i++) down=r[down];int pp=up;int dd=down;for(int i=0;i<h;i++){pp=d[pp]; dd=d[dd];swap(r[pp],r[dd]);}for(int i=0;i<w;i++){pp=r[pp];dd=r[dd];swap(d[pp],d[dd]);} pp=up; dd=down;  for(int i=0;i<w;i++){pp=r[pp];dd=r[dd];swap(d[pp],d[dd]);} for(int i=0;i<h;i++){pp=d[pp];dd=d[dd];swap(r[pp],r[dd]);}}int up=0,down;for(int i=1;i<=n;i++){   up=down=d[up];   for(int j=1;j<=m;j++)   {   down=r[down];   printf("%d ",a[down/(m+1)][down%(m+1)]);   }   puts("");   }return 0; }