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 = ;

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 = ;

    while (down--)

        x = d[x];

    while (right--)

        x = r[x];

    return x;

}

int main()

{

    scanf("%d%d%d", &n, &m, &q);

    for (int i = ; i <= n * m; i++)

    {

        scanf("%s", s + sum[i - ] + );

        sum[i] = sum[i - ] + strlen(s + sum[i - ] + );

    }

    for (int i = ; i <= n; i++)

        for (int j = ; j <= m; j++)

            a[i][j] = (i - ) * m + j;

    tot = n * m;

    for (int i = ; i <= n + ; i++)

        for (int j = ; j <= m + ; j++)

            if ((i || j) && !a[i][j])

                a[i][j] = ++tot;

    for (int i = ; i <= n; i++)

        for (int j = ; j <= m; j++)

        {

            r[a[i][j]] = a[i][j + ];

            d[a[i][j]] = a[i + ][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 - , y1 - ), Pos1 = d[r[pos1]];//先移动到方阵左上角的左上角一格

        int pos2 = pos(x2 - , y2 - ), Pos2 = d[r[pos2]];

        for (int i = , 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 = , 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 = ; i < c; i++) //跳到方阵最后一列

        {

            pos1 = r[pos1];

            pos2 = r[pos2];

        }

        for (int i = , 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 = ; i < l; i++) //跳到方阵最后一行

        {

            pos1 = d[pos1];

            pos2 = d[pos2];

        }

        for (int i = , 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 = , p1 = d[]; i <= n; i++, p1 = d[p1])

    {

        for (int j = , 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 - ] + ; k <= sum[p2]; k++)

                printf("%c", s[k]);

            printf(" ");

        }

        printf("\n");

    }

    return ;

}

巴特西

常州模拟赛d5t3 appoint

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