Transformations(USACO)

/* ID:tianlin2 PROG:transform LANG:C++ */ #include <iostream> #include <fstream> using namespace std; //相等 bool six(char a[][11],char c[][11],int h) { int i,j; for(i=0;i!=h;++i) for(j=0;j!=h;++j) { if(a[i][j]!=c[i][j]) return false; } return true; } //90° bool one(char a[][11],char c[][11],int h) { char b[11][11]; int n=0,i,j,x,y; while(n<h/2&&h%2==0 || n<=h/2&&h%2!=0) { i=n; j=n; while(j!=h-n) { y=h-n-1; x=j; b[x][y]=a[i][j]; ++j; } ++i; --j; while(i!=h-n) { x=h-n-1; y=h-1-i; b[x][y]=a[i][j]; ++i; } --i; while(j>=n) { y=n; x=j; b[x][y]=a[i][j]; --j; } ++j; --i; while(i>n) { x=n; y=h-1-i; b[x][y]=a[i][j]; --i; } ++n; } if(six(b,c,h)) return true; else return false; } //180° bool two(char a[][11],char c[][11],int h) { char b[11][11]; int n=0,i,j,x,y; while(n<h/2&&h%2==0 || n<=h/2&&h%2!=0) { i=n; j=n; while(j!=h-n) { x=h-n-1; y=h-1-j; b[x][y]=a[i][j]; ++j; } ++i; --j; while(i!=h-n) { x=h-1-i; y=n; b[x][y]=a[i][j]; ++i; } --i; while(j>=n) { x=n; y=h-1-j; b[x][y]=a[i][j]; --j; } ++j; --i; while(i>n) { x=h-1-i; y=h-n-1; b[x][y]=a[i][j]; --i; } ++n; } if(six(b,c,h)) return true; else return false; } //270° bool three(char a[][11],char c[][11],int h) { char b[11][11]; int n=0,i,j,x,y; while(n<h/2&&h%2==0 || n<=h/2&&h%2!=0) { i=n; j=n; while(j!=h-n) { x=h-1-j; y=n; b[x][y]=a[i][j]; ++j; } --j; ++i; while(i!=h-n) { x=n; y=i; b[x][y]=a[i][j]; ++i; } --i; while(j>=n) { x=h-1-j; y=h-n-1; b[x][y]=a[i][j]; --j; } ++j; --i; while(i>n) { x=h-1-n; y=i; b[x][y]=a[i][j]; --i; } ++n; } if(six(b,c,h)) return true; else return false; } //水平映射 bool four(char a[][11],char c[][11],int h) { char b[11][11]; int n=0,i,j,x,y; while(n<h/2&&h%2==0 || n<=h/2&&h%2!=0) { i=n; j=n; while(j!=h-n) { x=i; y=h-1-j; b[x][y]=a[i][j]; ++j; } ++i;--j; while(i!=h-n) { x=i; y=h-1-j; b[x][y]=a[i][j]; ++i; } --i; while(j>=n) { x=i; y=h-1-j; b[x][y]=a[i][j]; --j; } ++j;--i; while(i>n) { x=i; y=h-1-j; b[x][y]=a[i][j]; --i; } ++n; } if(six(b,c,h)) return true; else return false; } //混合 bool five(char a[][11],char c[][11],int h) { char b[11][11]; int n=0,i,j,x,y; while(n<h/2&&h%2==0 || n<=h/2&&h%2!=0) { i=n; j=n; while(j!=h-n) { x=i; y=h-1-j; b[x][y]=a[i][j]; ++j; } ++i;--j; while(i!=h-n) { x=i; y=h-1-j; b[x][y]=a[i][j]; ++i; } --i; while(j>=n) { x=i; y=h-1-j; b[x][y]=a[i][j]; --j; } ++j;--i; while(i>n) { x=i; y=h-1-j; b[x][y]=a[i][j]; --i; } ++n; } /*for(i=0;i!=h;++i) { for(j=0;j!=h;++j) { cout<<b[i][j]<<' '; } cout<<endl; }*/ if(one(b,c,h)) return true; else if(two(b,c,h)) return true; else if(three(b,c,h)) return true; else return false; } int main() { ofstream fout("transform.out"); ifstream fin("transform.in"); char a[11][11],b[11][11]; int n=0,i,j,x,y,h,c=0; fin>>h; for(i=0;i!=h;++i) for(j=0;j!=h;++j) fin>>a[i][j]; for(i=0;i!=h;++i) for(j=0;j!=h;++j) fin>>b[i][j]; if(one(a,b,h)) c=1; else if(two(a,b,h)) c=2; else if(three(a,b,h)) c=3; else if(four(a,b,h)) { c=4; } else if(five(a,b,h)) { c=5; } else if(six(a,b,h)) c=6; else c=7; fout<<c<<endl; return 0; }

主要利用分情况:1,2,3,4,5,6!每种情况又分矩阵的上下左右四种情况,很复杂!

其实像官方答案一样,上下左右的规律,都是一样的:

for(r=0; r<b.n; r++)
    for(c=0; c<b.n; c++)
        nb.b[c][b.n+1 - r] = b.b[r][c];

还有一点忽略了,180°就是90°再转90°,270°则是转三次90°,导致写多了两个函数!

还有没用到结构体!

官方答案:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

#define MAXN 10

typedef struct Board Board;
struct Board {
    int n;
    char b[MAXN][MAXN];
};
//90°的这个规律很好
/* rotate 90 degree clockwise: [r, c] -> [c, n+1 - r] */
Board
rotate(Board b)
{
    Board nb;
    int r, c;

    nb = b;
    for(r=0; r<b.n; r++)
    for(c=0; c<b.n; c++)
        nb.b[c][b.n+1 - r] = b.b[r][c];

    return nb;
}

/* reflect board horizontally: [r, c] -> [r, n-1 -c] */
Board
reflect(Board b)
{
    Board nb;
    int r, c;

    nb = b;
    for(r=0; r<b.n; r++)
    for(c=0; c<b.n; c++)
        nb.b[r][b.n-1 - c] = b.b[r][c];

    return nb;
}

/* return non-zero if and only if boards are equal */
int
eqboard(Board b, Board bb)
{
    int r, c;

    if(b.n != bb.n)
        return 0;

    for(r=0; r<b.n; r++)
    for(c=0; c<b.n; c++)
        if(b.b[r][c] != bb.b[r][c])
            return 0;
    return 1;
}

Board
rdboard(FILE *fin, int n)
{
    Board b;
    int r, c;

    b.n = n;
    for(r=0; r<n; r++) {
        for(c=0; c<n; c++)
            b.b[r][c] = getc(fin);
        assert(getc(fin) == '/n');
    }
    return b;
}

void
main(void)
{
    FILE *fin, *fout;
    Board b, nb;
    int n, change;

    fin = fopen("transform.in", "r");
    fout = fopen("transform.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d/n", &n);
    b = rdboard(fin, n);
    nb = rdboard(fin, n);
    //注意按照时针旋转的三种情况
    if(eqboard(nb, rotate(b)))
        change = 1;
    else if(eqboard(nb, rotate(rotate(b))))
        change = 2;
    else if(eqboard(nb, rotate(rotate(rotate(b)))))
        change = 3;
    else if(eqboard(nb, reflect(b)))
        change = 4;
    else if(eqboard(nb, rotate(reflect(b)))
         || eqboard(nb, rotate(rotate(reflect(b))))
         || eqboard(nb, rotate(rotate(rotate(reflect(b))))))
        change = 5;
    else if(eqboard(nb, b))
        change = 6;
    else
        change = 7;

    fprintf(fout, "%d/n", change);

    exit(0);
}