Transformations(USACO)

/*ID:tianlin2PROG:transformLANG: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 10typedef 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] */Boardrotate(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] */Boardreflect(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 */inteqboard(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;}Boardrdboard(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;}voidmain(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);}