这道题目甚长， 代码也是甚长， 但是思路却不是太难。然而有好多代码实现的细节， 确是十分的巧妙。 对代码阅读能力， 代码理解能力， 代码实现能力， 代码实现技巧， DFS方法都大有裨益， 敬请有兴趣者耐心细读。（也许由于博主太弱， 才有此等感觉）。
题目： UVa 1103
In order to understand early civilizations, archaeologists often study texts written in ancient languages. One such language, used in Egypt more than 3000 years ago, is based on characters called hieroglyphs. Figure C.1 shows six hieroglyphs and their names. In this problem, you will write a program to recognize these six characters.
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 < H200) 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.1. (Two figures are topologically equivalent if each can be transformed into the other by stretching without tearing.)
The last test case is followed by a line containing two zeros.
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
代码请细读（加了自己所理解的注释）：
1 // we pad one empty line/column to the top/bottom/left/right border, so color 1 is always "background" white 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<vector> 6 #include<set> 7 using namespace std; 8 9 char bin[256][5]; 10 11 const int maxh = 200 + 5; 12 const int maxw = 50 * 4 + 5; 13 14 int H, W, pic[maxh][maxw], color[maxh][maxw]; 15 char line[maxw]; 16 17 //把读入转化成 0， 1 的函数。 18 void decode(char ch, int row, int col) { 19 for(int i = 0; i < 4; i++) 20 pic[row][col+i] = bin[ch][i] - '0'; 21 } 22 23 const int dr[] = {-1, 1, 0, 0}; 24 const int dc[] = {0, 0, -1, 1}; 25 26 // dfs from (row, col) and paint color c 27 void dfs(int row, int col, int c) { 28 color[row][col] = c; 29 for(int i = 0; i < 4; i++) { 30 int row2 = row + dr[i]; 31 int col2 = col + dc[i]; 32 if(row2 >= 0 && row2 < H && col2 >= 0 && col2 < W && pic[row2][col2] == pic[row][col] && color[row2][col2] == 0) 33 dfs(row2, col2, c); 34 } 35 } 36 37 vector<set<int> > neighbors; 38 39 void check_neighbors(int row, int col) { 40 for(int i = 0; i < 4; i++) { 41 int row2 = row + dr[i]; 42 int col2 = col + dc[i]; 43 if(row2 >= 0 && row2 < H && col2 >= 0&& col2 < W && pic[row2][col2] == 0 && color[row2][col2] != 1) 44 neighbors[color[row][col]].insert(color[row2][col2]); 45 } 46 } 47 48 const char* code = "WAKJSD"; 49 //以容器长度来表示黑连通块旁的内白连通块数（即：象形文字内的白块数） 50 char recognize(int c) { 51 int cnt = neighbors[c].size(); 52 return code[cnt]; 53 } 54 55 // use this function to print the decoded picture 56 void print() { 57 for(int i = 0; i < H; i++) { 58 for(int j = 0; j < W; j++) printf("%d", pic[i][j]); 59 printf("\n"); 60 } 61 } 62 63 int main() { 64 strcpy(bin['0'], "0000");//用来转化成二进制， 此法甚妙 65 strcpy(bin['1'], "0001"); 66 strcpy(bin['2'], "0010"); 67 strcpy(bin['3'], "0011"); 68 strcpy(bin['4'], "0100"); 69 strcpy(bin['5'], "0101"); 70 strcpy(bin['6'], "0110"); 71 strcpy(bin['7'], "0111"); 72 strcpy(bin['8'], "1000"); 73 strcpy(bin['9'], "1001"); 74 strcpy(bin['a'], "1010"); 75 strcpy(bin['b'], "1011"); 76 strcpy(bin['c'], "1100"); 77 strcpy(bin['d'], "1101"); 78 strcpy(bin['e'], "1110"); 79 strcpy(bin['f'], "1111"); 80 81 int kase = 0; 82 while(scanf("%d%d", &H, &W) == 2 && H) { 83 memset(pic, 0, sizeof(pic)); 84 for(int i = 0; i < H; i++) { 85 scanf("%s", line); 86 for(int j = 0; j < W; j++) 87 decode(line[j], i+1, j*4+1);//把输入转发成二进制 存于数组pic[][]中 88 } 89 90 H += 2; 91 W = W * 4 + 2; 92 93 int cnt = 0; 94 vector<int> cc; // connected components of 1 95 memset(color, 0, sizeof(color));//标记数组。 96 for(int i = 0; i < H; i++) 97 for(int j = 0; j < W; j++) 98 if(!color[i][j]) { 99 dfs(i, j, ++cnt); 100 if(pic[i][j] == 1) cc.push_back(cnt);//扫描矩阵，且为所有连通块编号，并把黑连通块号存进cc容器中。 101 } 102 103 neighbors.clear(); 104 neighbors.resize(cnt+1);//设置容器大小， 不清楚的自行百度。嘿嘿！ 105 for(int i = 0; i < H; i++) 106 for(int j = 0; j < W; j++) 107 if(pic[i][j] == 1) 108 check_neighbors(i, j);//扫描黑点，并把该点旁边有几个内连通白块存入neighbors容器。//其中下标为黑点对应的连通块编号。 109 110 vector<char> ans; 111 for(int i = 0; i < cc.size(); i++) 112 ans.push_back(recognize(cc[i]));//存目标值， 并排序。 113 sort(ans.begin(), ans.end()); 114 115 printf("Case %d: ", ++kase); 116 for(int i = 0; i < ans.size(); i++) printf("%c", ans[i]); 117 printf("\n"); 118 } 119 return 0; 120 }