CF367 E - Working routine

十字链表模拟

#include<bits/stdc++.h>
using namespace std;

int n,m,q;
struct Node{
	int v; int d,r;
}ma[1005*1005];
int C(int x,int y){
	return x*(m+1)+y;
}

int main(){
	while(~scanf("%d %d %d",&n,&m,&q)) {
		for(int i = 1; i <= n; ++i) 
			for(int j = 1; j <= m; ++j) {
				scanf("%d",&ma[C(i,j)].v);
			}
		for(int i = 0; i <= n; ++i)
			for(int j = 0; j <= m; ++j) {
				ma[C(i,j)].r = C(i,j+1);
				ma[C(i,j)].d = C(i+1,j);
			}
		for(int i = 0; i < q; ++i) {
			int a,b,c,d,h,w; scanf("%d %d %d %d %d %d",&a,&b,&c,&d,&h,&w);
			int t1 = 0; int t2 = 0;
			for(int i = 1; i < a; ++i) t1 = ma[t1].d; 
			for(int i = 1; i < b; ++i) t1 = ma[t1].r;
			for(int i = 1; i < c; ++i) t2 = ma[t2].d;
			for(int i = 1; i < d; ++i) t2 = ma[t2].r;
			int x1,x2;
			
			x1 = t1; x2 = t2; 
			for(int i = 1; i <= w; ++i) { 
				x1 = ma[x1].r; x2 = ma[x2].r;
				swap(ma[x1].d, ma[x2].d); 
			}
			for(int i = 1; i <= h; ++i) {
				x1 = ma[x1].d; x2 = ma[x2].d;
				swap(ma[x1].r, ma[x2].r);
			}

			x1 = t1, x2 = t2;
			for(int i = 1; i <= h; ++i) {
				x1 = ma[x1].d; x2 = ma[x2].d;
				swap(ma[x1].r, ma[x2].r);
			}
			for(int i = 1; i <= w; ++i) {
				x1 = ma[x1].r; x2 = ma[x2].r;
				swap(ma[x1].d, ma[x2].d);
			}
		}

		int tt = 0; 
		for(int i = 1; i <= n; ++i) {
			tt = ma[tt].d;
			int t1 = tt;
			for(int j = 1; j <= m; ++j) {
				t1 = ma[t1].r; 
				if(j!=1) printf(" ");
				printf("%d",ma[t1].v);		
			} printf("\n");
		}

	}
	return 0;
}