Flip and Turn
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 957 | Accepted: 330 |
Description
Let us define a set of operations on a rectangular matrix of printable characters.
A matrix A with m rows (1-st index) and n columns (2-nd index) is given. The resulting matrix B is defined as follows.
Transposition by the main diagonal (operation identifier is ‘1’): Bj,i =Ai,jTransposition by the second diagonal (‘2’): Bn?j+1,m?i+1 = Ai,jHorizontal flip (‘H’): Bm?i+1,j = Ai,jVertical flip (‘V’): Bi,n?j+1 = Ai,jRotation by 90 (‘A’), 180 (‘B’), or 270 (‘C’) degrees clockwise; 90 degrees case: Bj,m?i+1 = Ai,jRotation by 90 (‘X’), 180 (‘Y’), or 270 (‘Z’) degrees counterclockwise; 90 degrees case: Bn?j+1,i =Ai,j
You are given a sequence of no more than 100 000 operations from the set. Apply the operations to the given matrix and output the resulting matrix.
Input
At the first line of the input file there are two integer numbers — m and n (0 < m, n≤ 300). Then there are m lines with n printable characters per line (we define a printable character as a symbol with ASCII code from 33 to 126 inclusive). There will be no additional symbols at these lines.
The next line contains the sequence operations to be performed, specified by their one-character identifiers. The operations should be performed from left to right.
Output
Two integer numbers, the number of rows and columns in the output matrix. Then the output matrix must follow, in the same format as the input one.
Sample Input
1 2 3 4 5 |
3 4 0000 a0b0 cdef A1 |
Sample Output
1 2 3 4 |
3 4 cdef a0b0 0000 |
题目大意：给你一个m*n的矩阵，通过顺时针旋转，逆时针旋转，水平翻转，垂直翻转，对角线旋转，反对角线旋转变换得到新矩阵。但是不能直接对原矩阵操作，时间复杂度为O(10^5*10^5)，会超时，但是如何保存变化的过程呢，我们不需要知道变化过程，只需要知道结果即可。由于旋转翻转都是中心对称的，可以采用一个2*2的小正方形记录变化情况，然后小矩阵变化得出最终结果之后再相应变化大矩阵。时间复杂度为O(10^5*5)
题目地址：Flip and Turn
AC代码：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 |
#include #include #include #include #include #include using namespace std; char a[305][305]; char b[305][305]; char str[100005]; int m,n; int tb[3][3]; void con1() { int i,j; for(i=1; i<=n; i++) { for(j=1; j<=m; j++) { b[i][j]=a[j][i]; } } for(i=1; i<=n; i++) { for(j=1; j<=m; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } swap(n,m); } void con2() { int i,j; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { b[n-j+1][m-i+1]=a[i][j]; } } for(i=1; i<=n; i++) { for(j=1; j<=m; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } swap(n,m); } void con3() { int i,j; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { b[m-i+1][j]=a[i][j]; } } for(i=1; i<=m; i++) { for(j=1; j<=n; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } } void con4() { int i,j; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { b[i][n-j+1]=a[i][j]; } } for(i=1; i<=m; i++) { for(j=1; j<=n; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } } void con5() { int i,j; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { b[j][m-i+1]=a[i][j]; } } for(i=1; i<=n; i++) { for(j=1; j<=m; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } swap(m,n); } void con6() { int i,j; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { b[n-j+1][i]=a[i][j]; } } for(i=1; i<=n; i++) { for(j=1; j<=m; j++) a[i][j]=b[i][j]; a[i][j]='\0'; } swap(m,n); } void debug() { int i,j; for(i=1;i<=2;i++) { for(j=1;j<=2;j++) cout< { int s=str[i]-'A'+1; for(j=0;j { int s=str[i]-'X'+1; for(j=0;j
|