常州模拟赛d5t3 appoint

分析：这道题比较奇葩.因为字符串没有swap函数，所以一个一个字符串交换只有30分.但是我们可以不用直接交换字符串，而是交换字符串的指针，相当于当前位置是哪一个字符串，每次交换int,可以拿60分.

      对于二维问题，通常转化为一维问题去考虑，得到适当的方法再应用到二维上来，这道题如果转移到一维上就是给你一个序列，每次交换一对区间，区间不重叠，最后要求顺序输出整个序列，很显然，我们只要记录每个数旁边的数就好了，所以用链表能很快解决.转化到二维上，我们记录一个右方的链表，下方的链表，每次交换操作只需要更改四周的链表就好了.

      需要注意的是char数组不能够开成2维的，题目中只告诉了字符串的总长度，因此需要转化为一维的,输出则在前一个字符串的基础上输出.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1010;

int n, m, q, a[maxn][maxn], sum[maxn * maxn], r[maxn * maxn], d[maxn * maxn],tot;
char s[maxn * maxn];

int pos(int down, int right)
{
    int x = 0;
    while (down--)
        x = d[x];
    while (right--)
        x = r[x];
    return x;
}

int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n * m; i++)
    {
        scanf("%s", s + sum[i - 1] + 1);
        sum[i] = sum[i - 1] + strlen(s + sum[i - 1] + 1);
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            a[i][j] = (i - 1) * m + j;
    tot = n * m;
    for (int i = 0; i <= n + 1; i++)
        for (int j = 0; j <= m + 1; j++)
            if ((i || j) && !a[i][j])
                a[i][j] = ++tot;
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= m; j++)
        {
            r[a[i][j]] = a[i][j + 1];
            d[a[i][j]] = a[i + 1][j];
        }
    while (q--)
    {
        int x1, y1, x2, y2, l, c;
        scanf("%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &l, &c);
        int pos1 = pos(x1 - 1, y1 - 1), Pos1 = d[r[pos1]];//先移动到方阵左上角的左上角一格
        int pos2 = pos(x2 - 1, y2 - 1), Pos2 = d[r[pos2]];
        for (int i = 1, p1 = d[pos1], p2 = d[pos2]; i <= l; i++, p1 = d[p1], p2 = d[p2]) //更改方阵第一列左边一列
            swap(r[p1], r[p2]);
        for (int i = 1, p1 = r[pos1], p2 = r[pos2]; i <= c; i++, p1 = r[p1], p2 = r[p2])
            swap(d[p1], d[p2]);
        pos1 = Pos1, pos2 = Pos2;
        for (int i = 1; i < c; i++) //跳到方阵最后一列
        {
            pos1 = r[pos1];
            pos2 = r[pos2];
        }
        for (int i = 1, p1 = pos1, p2 = pos2; i <= l; i++, p1 = d[p1], p2 = d[p2]) //交换方阵最后一列
            swap(r[p1], r[p2]);
        pos1 = Pos1, pos2 = Pos2;
        for (int i = 1; i < l; i++) //跳到方阵最后一行
        {
            pos1 = d[pos1];
            pos2 = d[pos2];
        }
        for (int i = 1, p1 = pos1, p2 = pos2; i <= c; i++, p1 = r[p1], p2 = r[p2]) //交换方阵最后一行
            swap(d[p1], d[p2]);
    }
    for (int i = 1, p1 = d[0]; i <= n; i++, p1 = d[p1])
    {
        for (int j = 1, p2 = r[p1]; j <= m; j++, p2 = r[p2]) //p2千万不能写成r[0],有可能d[0]和r[0]不是同一格
        {
            for (int k = sum[p2 - 1] + 1; k <= sum[p2]; k++)
                printf("%c", s[k]);
            printf(" ");
        }
        printf("
");
    }

    return 0;
}