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文章作者：濟(jì)南軟件開發(fā) 時間：2016年12月20日
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 |
題目大意：給你一個m*n的矩陣，通過順時針旋轉(zhuǎn)，逆時針旋轉(zhuǎn)，水平翻轉(zhuǎn)，垂直翻轉(zhuǎn)，對角線旋轉(zhuǎn)，反對角線旋轉(zhuǎn)變換得到新矩陣。但是不能直接對原矩陣操作，時間復(fù)雜度為O(10^5*10^5)，會超時，但是如何保存變化的過程呢，我們不需要知道變化過程，只需要知道結(jié)果即可。由于旋轉(zhuǎn)翻轉(zhuǎn)都是中心對稱的，可以采用一個2*2的小正方形記錄變化情況，然后小矩陣變化得出最終結(jié)果之后再相應(yīng)變化大矩陣。時間復(fù)雜度為O(10^5*5)
題目地址：Flip and Turn
AC代碼：
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<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #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<<tb[i][j]<<" ";="" cout<<endl;="" }="" int="" main()="" {="" i,j;="" while(~scanf("%d%d",&m,&n))="" for(i="1;" i<="m;" i++)="" scanf("%s",a[i]+1);="" scanf("%s",str);="" tb[1][1]="1,tb[1][2]=2,tb[2][1]=3,tb[2][2]=4;" i<strlen(str);="" if(str[i]="='1')" swap(tb[1][2],tb[2][1]);="" else="" swap(tb[1][1],tb[2][2]);="" swap(tb[1][1],tb[2][1]);="" swap(tb[1][2],tb[2][2]);="" swap(tb[1][1],tb[1][2]);="" swap(tb[2][1],tb[2][2]);="">='A'&&str[i]<='C') { int s=str[i]-'A'+1; for(j=0;j<s;j++) {="" swap(tb[2][2],tb[1][2]);="" swap(tb[1][1],tb[1][2]);="" swap(tb[1][1],tb[2][1]);="" }="" else="" if(str[i]="">='X'&&str[i]<='Z') { int s=str[i]-'X'+1; for(j=0;j<s;j++) {="" swap(tb[1][1],tb[2][1]);="" swap(tb[1][1],tb[1][2]);="" swap(tb[2][2],tb[1][2]);="" }="" debug();="" 是中心對稱，順時針轉(zhuǎn)，對稱轉(zhuǎn)，沿對角線折都是中心對稱，所以1="" 4只能是對角="" if(tb[1][1]="=1&&tb[1][2]==2&&tb[2][1]==3&&tb[2][2]==4)" {}="" 1="" 2="" 3="" 4="" else="" con1();="" con4();="" con5();="" con2();="" {con2();="" con1();}="" con6();="" {con6();="" con2();}="" cout<<m<<"="" "<<n<<endl;="" for(i="1;" i<="m;" i++)="" cout<<br> <br>
<br> </s;j++)></s;j++)></tb[i][j]<<"></algorith |
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