这道搜索题没什么绕弯的地方，把transformation用矩阵位置变换表示，然后一一验证即可。
对于#5 combination我写得比较啰嗦: )
/* ID: alexyua2 PROG: transform LANG: C++ */ #include<fstream> using namespace std; ifstream fin("transform.in"); ofstream fout("transform.out"); //ifstream fin("in.txt"); //ofstream fout("out.txt"); int MAX = 11; int main() { //input int N; fin>>N; char before[MAX][MAX],after[MAX][MAX]; int i,j; for(i=1;i<=N;i++) for(j=1;j<=N;j++) fin>>before[i][j]; for(i=1;i<=N;i++) for(j=1;j<=N;j++) fin>>after[i][j]; //search transformation pattern //#1:90 Degree Rotation int state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(before[i][j] != after[j][N+1-i]) { state = 0; break; } if(state == 1) { fout<<"1" <<endl; return 0; } //#2:180 Degree Rotation state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(before[i][j] != after[N+1-i][N+1-j]) { state = 0; break; } if(state == 1) { fout<<"2" <<endl; return 0; } //#3:270 Degree Rotation state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(before[i][j] != after[N+1-j][i]) { state = 0; break; } if(state == 1) { fout<<"3" <<endl; return 0; } //#4:Reflection state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(before[i][j] != after[i][N+1-j]) { state = 0; break; } if(state == 1) { fout<<"4" <<endl; return 0; } //#5:Combination char temp[MAX][MAX]; for(i=1;i<=N;i++) for(j=1;j<=N;j++) temp[i][j] = before[i][N+1-j]; state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(temp[i][j] != after[j][N+1-i]) { state = 0; break; } if(state == 1) { fout<<"5" <<endl; return 0; } state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(temp[i][j] != after[N+1-i][N+1-j]) { state = 0; break; } if(state == 1) { fout<<"5" <<endl; return 0; } state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(temp[i][j] != after[N+1-j][i]) { state = 0; break; } if(state == 1) { fout<<"5" <<endl; return 0; } //#6:No Change state = 1; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if(before[i][j] != after[i][j]) { state = 0; break; } if(state == 1) { fout<<"6" <<endl; return 0; } //#7:No Answer fout<<"7" <<endl; //close files fin.close(); fout.close(); return 0; }