Codeforces Round #367 (Div. 2) E. Working routine (十字链表，二维高效交换）

分析

链表和数组是两对极端，链表定位慢，增删交换快。而块状链表是二者的综合体
这里大概是一维链表高效交换的二维版本。
维护一个十字链表即可。
十字链表没有必要记录每个元素上下左右的所有元素，只需要记录下和右即可，维持要增加第0行和0列。
对于矩阵交换，暴力的更新整个矩阵周长两边的节点的值即可，复杂度$O(q \times (m + n)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define pr(x) cout << #x << ": " << x << "  " 
#define pl(x) cout << #x << ": " << x << endl;

struct node {
    int v, r, d;
    node (){}
    node (int V, int R, int D): v(V), r(R), d(D){}
} nodes[1009 * 1009];


struct jibancanyang
{
    int n, m, q;

    int f(int i, int j) {
        return i * (m + 1) + j;
    }
    int g(int i, int j) {
        int k = 0;
        while (i--) k = nodes[k].d;
        while (j--) k = nodes[k].r;
        return k;
    }

    void print() {
        int k = 0;
        for (int i = 0; i < n; ++i) {
            k = nodes[k].d;
            for (int j = 0, t = k; j < m; ++j) {
                t = nodes[t].r;
                printf("%d ", nodes[t].v);
            }
            puts("");
        }
    }

    void build() {
        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= m; ++j) {
                nodes[f(i, j)] = node(0, f(i, j + 1), f(i + 1, j));
            }
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                scanf("%d", &nodes[f(i, j)].v);
            }
        }
        while (q--) {
            int a, b, x, y, l, w;
            scanf("%d%d%d%d%d%d", &a, &b, &x, &y, &w, &l);
            a = g(a - 1, b - 1), b = g(x - 1, y - 1);
//            pr(nodes[a].v), pr(nodes[a].r), pl(nodes[a].d);
            x = a, y = b;
            for (int i = 0; i < l; ++i) {
                a = nodes[a].r, b = nodes[b].r;
                swap(nodes[a].d, nodes[b].d);
            }
            for (int i = 0; i < w; ++i) {
                a = nodes[a].d, b = nodes[b].d;
                swap(nodes[a].r, nodes[b].r);
            }

            for (int i = 0; i < w; ++i) {
                x = nodes[x].d, y = nodes[y].d;
                swap(nodes[x].r, nodes[y].r);
            }
            for (int i = 0; i < l; ++i) {
                x = nodes[x].r, y = nodes[y].r;
                swap(nodes[x].d, nodes[y].d);
            }
        }
        print();
    }
}ac;

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    scanf("%d%d%d", &ac.n, &ac.m, &ac.q);
    ac.build();
    return 0;
}