Input
The input consists of several test cases, each of which describes an image containing one or more hieroglyphs chosen from among those shown in Figure C.1. The image is given in the form of a series of horizontal scan lines consisting of black pixels (represented by 1) and white pixels (represented by 0). In the input data, each scan line is encoded in hexadecimal notation. For example, the sequence of eight pixels 10011100 (one black pixel, followed by two white pixels, and so on) would be represented in hexadecimal notation as 9c. Only digits and lowercase letters a through f are used in the hexadecimal encoding. The first line of each test case contains two integers, H and W: H (0 < H <= 200) is the number of scan lines in the image. W (0 < W <= 50) is the number of hexadecimal characters in each line. The next H lines contain the hexadecimal characters of the image, working from top to bottom. Input images conform to the following rules:- The image contains only hieroglyphs shown in Figure C.1.
- Each image contains at least one valid hieroglyph.
- Each black pixel in the image is part of a valid hieroglyph.
- Each hieroglyph consists of a connected set of black pixels and each black pixel has at least one other black pixel on its top, bottom, left, or right side.
- The hieroglyphs do not touch and no hieroglyph is inside another hieroglyph.
- Two black pixels that touch diagonally will always have a common touching black pixel.
- The hieroglyphs may be distorted but each has a shape that is topologically equivalent to one of the symbols in Figure C.11.
The last test case is followed by a line containing two zeros.
1Two figures are topologically equivalent if each can be transformed into the other by stretching without tearing.
Output
For each test case, display its case number followed by a string containing one character for each hieroglyph recognized in the image, using the following code:Ankh: A
Wedjat: J
Djed: D
Scarab: S
Was: W
Akhet: K
In each output string, print the codes in alphabetic order. Follow the format of the sample output.
The sample input contains descriptions of test cases shown in Figures C.2 and C.3. Due to space constraints not all of the sample input can be shown on this page.
Sample Input
100 25 0000000000000000000000000 0000000000000000000000000 ...(50 lines omitted)... 00001fe0000000000007c0000 00003fe0000000000007c0000 ...(44 lines omitted)... 0000000000000000000000000 0000000000000000000000000 150 38 00000000000000000000000000000000000000 00000000000000000000000000000000000000 ...(75 lines omitted)... 0000000003fffffffffffffffff00000000000 0000000003fffffffffffffffff00000000000 ...(69 lines omitted)... 00000000000000000000000000000000000000 00000000000000000000000000000000000000 0 0
Sample Output
Case 1: AKW Case 2: AAAAA
题意:给出十六进制的数字矩阵,对其转换为二进制矩阵,输出转换后图像中按字典序排列的象形文字名。
解题思路:给出的象形文字的中间洞数各不相同(那就可以操作一波了,那么只要先找到字,在找出该字内部有几个洞即可确定象形字母的类型,首先去除字外的0,将其标为-1,
这里有个小操作是把存的图像外圈围一圈0,这样就可以直接从(0,0)点dfs对字外的0改标记为1,然后再遍历数组找1,即找字,找到字后将字内的0同样标上-1,并计算空洞数,
同时将字上的1全部转换(相当于遍历过这里的标记),确定这个字之后将其压入vector数组,最后结果对vector数组排序并输出即可。
1 #include <algorithm> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstring> 5 #include <iostream> 6 #include <cstdlib> 7 #include <set> 8 #include <vector> 9 #include <cctype> 10 #include <iomanip> 11 #include <sstream> 12 #include <climits> 13 #include <queue> 14 #include <stack> 15 using namespace std; 16 typedef long long ll; 17 #define INF 0X3f3f3f3f 18 const ll MAXN = 1e5 + 7; 19 const ll MOD = 1e9 + 7; 20 char trans[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'}; 21 int transto[][4] = { 22 {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 1, 1}, 23 {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 1, 1, 0}, {0, 1, 1, 1}, 24 {1, 0, 0, 0}, {1, 0, 0, 1}, {1, 0, 1, 0}, {1, 0, 1, 1}, 25 {1, 1, 0, 0}, {1, 1, 0, 1}, {1, 1, 1, 0}, {1, 1, 1, 1} 26 }; 27 int maze[500][500]; 28 int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; 29 char ans[] = {'W', 'A', 'K', 'J', 'S', 'D'}; 30 int n, m, cnt=0; 31 bool check(int x, int y) 32 { 33 if (x >= 0 && x <= n+1 && y >= 0 && y <= 1+4 * m) 34 return true; 35 return false; 36 } 37 void dfs_vis(int x, int y) //将字外0标为-1 38 { 39 if (!check(x, y)||maze[x][y]) 40 return; 41 maze[x][y] = -1; //在字外面的0标为-1 42 for (int i = 0; i < 4; i++) 43 dfs_vis(x + dir[i][0], y + dir[i][1]); 44 } 45 void dfs_bak(int x, int y) //找字的空白块 46 { 47 if (!check(x, y) ||maze[x][y] == -1 || maze[x][y] == 2) 48 return; 49 if (maze[x][y] == 0) 50 { 51 cnt++; 52 dfs_vis(x, y); 53 } 54 else 55 { 56 maze[x][y] = 2; 57 for (int i = 0; i < 4; i++) 58 dfs_bak(x + dir[i][0], y + dir[i][1]); 59 } 60 return; 61 } 62 void debug() 63 { 64 for (int i = 0; i <= n+1; i++) 65 { 66 for (int j = 0; j <=1+ 4 * m; j++) 67 cout << maze[i][j]; 68 cout << endl; 69 } 70 return; 71 } 72 int main() 73 { 74 ios::sync_with_stdio(false); 75 int t = 1; 76 while (cin >> n >> m && n && m) 77 { 78 vector<char> vec; 79 memset(maze, 0, sizeof(maze)); 80 for (int i = 0; i < n; i++) 81 { 82 char temp[60]; 83 cin >> temp; 84 int cnt = 1; 85 for (int i1 = 0; i1 < m; i1++) 86 { 87 for (int j = 0; j < 16; j++) 88 { 89 if (trans[j] == temp[i1]) 90 { 91 for (int k = 0; k < 4; k++) 92 maze[i+1][cnt++] = transto[j][k]; 93 break; 94 } 95 } 96 } 97 } 98 // debug(); 99 dfs_vis(0,0); 100 for (int i = 1; i <= n; i++) 101 { 102 for (int j = 1; j <= 4*m+1; j++) 103 { 104 if (maze[i][j] == 1) 105 { 106 dfs_bak(i, j); 107 vec.push_back(ans[cnt]); 108 cnt=0; 109 // debug(); 110 } 111 } 112 } 113 sort(vec.begin(), vec.end()); 114 cout << "Case " << t++ << ": "; 115 vector<char>::iterator it; 116 for (it = vec.begin(); it != vec.end(); it++) 117 cout << *it; 118 cout << endl; 119 } 120 return 0; 121 } 122 /* A-1 J-3 D-5 S-4 W-0 K-2*/
再给出两组数据方便自行debug:
100 25
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150 38
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Case 1: AKW Case 2: AAAAA