USACO section1.2 Transformations

模拟题，给出一个 n×n 的字符块的两个状态（分别为初始态和目标态）和几种操作，问目标态最少是由初始态的哪个操作完成的，WA 3 次。

/*
PROG: transform
LANG: C++
*/
# include <cstdio>
# include <cstring>

# define N 10 + 5

void print(char s[][N], int n)
{
    for (int i = 0; i < n; ++i)
        puts(s[i]);
}

/* rotate 90 degrees clockwise */
void rotate(char s[][N], int n)
{
    char buf[N][N];

    memset(buf, 0, sizeof(buf));
    for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
        buf[i][j] = s[n-j-1][i];
    memcpy(s, buf, sizeof(buf));
    //print(s, n);
}

void mirror(char s[][N], int n)
{
    char buf[N][N];

    memset(buf, 0, sizeof(buf));
    for (int i = 0; i < n; /*puts(buf[i]), */++i)
    for (int j = 0; j < n; ++j)
    {
        buf[i][j] = s[i][n-1-j];
    }
    memcpy(s, buf, sizeof(buf));
   // print(s, n);
}

char isEqual(char s[][N], char g[][N], int n)
{
    for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
        if (s[i][j] != g[i][j]) return 0;
    return 1;
}

int main()
{
    freopen("transform.in", "r", stdin);
    freopen("transform.out", "w", stdout);

    int n;
    char s[N][N], f[N][N], g[N][N];

    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
        scanf("%s", s[i]);
    for(int i = 0; i < n; ++i)
        scanf("%s", g[i]);

    memcpy(f, s, sizeof(s));
    if (rotate(f, n), isEqual(f, g, n)) printf("1\n");
    else if (rotate(f, n), isEqual(f, g, n)) printf("2\n");
    else if (rotate(f, n), isEqual(f, g, n)) printf("3\n");
    else if (rotate(f, n), isEqual(s, g, n)) printf("6\n");
    else if (mirror(f, n), isEqual(f, g, n)) printf("4\n");
    else if (rotate(f, n), isEqual(f, g, n)) printf("5\n");
    else if (rotate(f, n), isEqual(f, g, n)) printf("5\n");
    else if (rotate(f, n), isEqual(f, g, n)) printf("5\n");
    else printf("7\n");

    fclose(stdin);
    fclose(stdout);

    return 0;
}

/**/