Codeforces Round #367 (Div. 2)E

题意：

给定一个矩阵，每次操作给出两个子矩阵参数，交换这两个子矩阵，

子矩阵满足：

不相交，没有公共边

要求最后输出这个矩阵

题解：构建四向链表，存每个点上面下面左边右边的点，每次修改就先通过链表找出每个子矩阵的左上角，然后

四条边去瞎搞一通O(N)

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<bitset>
#include<algorithm>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<cmath>
using namespace std;

const int maxn = 1010;

int n,m,q,num[maxn][maxn],x,y,pos[maxn][maxn],
le[maxn*maxn + 4*maxn],ri[maxn*maxn + 4*maxn],
up[maxn*maxn + 4*maxn],dw[maxn*maxn + 4*maxn];

int getpos(int a,int b)
{
int now = pos[a][0];
for (int i = 1; i <= b; i++) now = ri[now];
return now;
}

int Change(int now)
{
x = now/m + 1; if (now % m == 0) --x;
y = now%m; if (!y) y = m;
}

int main()
{
#ifdef DMC
freopen("DMC.txt","r",stdin);
#endif

cin >> n >> m >> q;
int tot = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) scanf("%d",&num[i][j]),pos[i][j] = ++tot;
for (int i = 1; i <= n; i++) {
pos[i][0] = ++tot; pos[i][m+1] = ++tot;
for (int j = 1; j <= m; j++) {
le[pos[i][j]] = pos[i][j-1];
ri[pos[i][j]] = pos[i][j+1];
up[pos[i][j]] = pos[i-1][j];
dw[pos[i][j]] = pos[i+1][j];
}
ri[pos[i][0]] = pos[i][1];
le[pos[i][m+1]] = pos[i][m];
}
for (int i = 1; i <= m; i++) {
pos[0][i] = ++tot; pos[n+1][i] = ++tot;
dw[pos[0][i]] = pos[1][i]; up[pos[1][i]] = pos[0][i];
up[pos[n+1][i]] = pos
[i]; dw[pos
[i]] = pos[n+1][i];
}

while (q--) {
int A1,B1,A2,B2,h,w;
scanf("%d%d%d%d%d%d",&A1,&B1,&A2,&B2,&h,&w);
int posA = getpos(A1,B1);
int posB = getpos(A2,B2);
for (int i = A1,j = A2; i < A1 + h; i++,j++) {
swap(ri[le[posA]],ri[le[posB]]);
swap(le[posA],le[posB]);
posA = dw[posA]; posB = dw[posB];
}
posA = up[posA]; posB = up[posB];
for (int i = B1,j = B2; i < B1 + w; i++,j++) {
swap(up[dw[posA]],up[dw[posB]]);
swap(dw[posA],dw[posB]);
posA = ri[posA]; posB = ri[posB];
}
posA = le[posA]; posB = le[posB];
for (int i = A1 + h - 1,j = A2 + h - 1; i >= A1; i--,j--) {
swap(le[ri[posA]],le[ri[posB]]);
swap(ri[posA],ri[posB]);
posA = up[posA]; posB = up[posB];
}
posA = dw[posA]; posB = dw[posB];
for (int i = B1 + w - 1, j = B2 + w - 1; i >= B1; i--,j--) {
swap(dw[up[posA]],dw[up[posB]]);
swap(up[posA],up[posB]);
posA = le[posA]; posB = le[posB];
}
}
4000

for (int i = 1; i <= n; i++) {
int now = pos[i][0];
for (int j = 1; j <= m; j++) {
now = ri[now];
Change(now);
printf("%d ",num[x][y]);
}
printf("\n");
}
return 0;
}