常州類比賽d5t3 appoint

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標籤:基礎   include   code   個數   wap   down   i++   分享   數組   

分析:這道題比較奇葩.因為字串沒有swap函數,所以一個一個字串交換隻有30分.但是我們可以不用直接交換字串,而是交換字串的指標,相當於當前位置是哪一個字串,每次交換int,可以拿60分.

      對於二維問題,通常轉化為一維問題去考慮,得到適當的方法再應用到二維上來,這道題如果轉移到一維上就是給你一個序列,每次交換一對區間,區間不重疊,最後要求順序輸出整個序列,很顯然,我們只要記錄每個數旁邊的數就好了,所以用鏈表能很快解決.轉化到二維上,我們記錄一個右方的鏈表,下方的鏈表,每次交換操作只需要更改四周的鏈表就好了.

      需要注意的是char數組不能夠開成2維的,題目中只告訴了字串的總長度,因此需要轉化為一維的,輸出則在前一個字串的基礎上輸出.

#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]];//先移動到方陣左上方的左上方一格        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]) //更改方陣第一列左邊一列            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++) //跳到方陣最後一列        {            pos1 = r[pos1];            pos2 = r[pos2];        }        for (int i = 1, p1 = pos1, p2 = pos2; i <= l; i++, p1 = d[p1], p2 = d[p2]) //交換方陣最後一列            swap(r[p1], r[p2]);        pos1 = Pos1, pos2 = Pos2;        for (int i = 1; i < l; i++) //跳到方陣最後一行        {            pos1 = d[pos1];            pos2 = d[pos2];        }        for (int i = 1, p1 = pos1, p2 = pos2; i <= c; i++, p1 = r[p1], p2 = r[p2]) //交換方陣最後一行            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]) //p2千萬不能寫成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("\n");    }    return 0;}

 

常州類比賽d5t3 appoint

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