图像被压缩过，所以先解压成01矩阵，把所有的连通块填充颜色，背景色为1，黑色连通块的标号存放在cc中。neighbors是存放白洞的数组，最后根据白洞来判断属于那个字符。
AC代码：
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; const int maxh = 205; const int maxw = 205; char bin[256][5]; int H, W, pic[maxh][maxw], color[maxh][maxw]; char line[maxw]; // 解码成一个01矩阵 void decode(char ch, int row, int col) { for (int i = 0; i < 4; i++) { pic[row][col + i] = bin[ch][i] - '0'; } } const int dr[] = { -1, 1, 0, 0 }; const int dc[] = { 0, 0, -1, 1 }; // DFS遍历，把连通块标记为C void dfs(int row, int col, int c) { color[row][col] = c; for (int i = 0; i < 4; i++) { // 这道题只需要考虑上下左右 int row2 = row + dr[i]; int col2 = col + dc[i]; if (row2 >= 0 && row2 < H && col2 >= 0 && col2 < W && pic[row2][col2] == pic[row][col] && color[row2][col2] == 0) { dfs(row2, col2, c); } } } // 集合数组，记录黑色连通块周围相连的非背景的白块，也就是白洞 vector > neighbors; void checkNeighbors(int row, int col) { for (int i = 0; i < 4; i++) { int row2 = row + dr[i]; int col2 = col + dc[i]; if (row2 >= 0 && row2 < H && col2 >= 0 && col2 < W && pic[row2][col2] == 0 && color[row2][col2] != 1) { neighbors[color[row][col]].insert(color[row2][col2]); } } } const char* code = "WAKJSD"; char recognize(int c) { int cnt = neighbors[c].size(); return code[cnt]; } void print() { for (int i = 0; i < H; i++) { for (int j = 0; j < W; j++) { printf("%d", pic[i][j]); } printf("\n"); } } int main() { strcpy(bin['0'], "0000"); strcpy(bin['1'], "0001"); strcpy(bin['2'], "0010"); strcpy(bin['3'], "0011"); strcpy(bin['4'], "0100"); strcpy(bin['5'], "0101"); strcpy(bin['6'], "0110"); strcpy(bin['7'], "0111"); strcpy(bin['8'], "1000"); strcpy(bin['9'], "1001"); strcpy(bin['a'], "1010"); strcpy(bin['b'], "1011"); strcpy(bin['c'], "1100"); strcpy(bin['d'], "1101"); strcpy(bin['e'], "1110"); strcpy(bin['f'], "1111"); int kase = 0; while (scanf("%d%d", &H, &W) == 2 && H) { memset(pic, 0, sizeof(pic)); for (int i = 0; i < H; i++) { scanf("%s", line); for (int j = 0; j < W; j++) { decode(line[j], i + 1, j * 4 + 1); } } H += 2; W = W * 4 + 2; int cnt = 0; vector cc; // 黑色连通块的标号存放在cc里面 memset(color, 0, sizeof(color)); for (int i = 0; i < H; i++) { for (int j = 0; j < W; j++) { if (!color[i][j]) { dfs(i, j, ++cnt); if (pic[i][j] == 1) { cc.push_back(cnt); } } } } neighbors.clear(); neighbors.resize(cnt + 1); for (int i = 0; i < H; i++) { for (int j = 0; j < W; j++) { if (pic[i][j] == 1) { checkNeighbors(i, j); } } } vector ans; for (int i = 0; i < cc.size(); i++) { ans.push_back(recognize(cc[i])); } sort(ans.begin(), ans.end()); printf("Case %d: ", ++kase); for (int i = 0; i < ans.size(); i++) { printf("%c", ans[i]); } printf("\n"); } return 0; }
转载于:https://www.cnblogs.com/zhangyaoqi/p/4591559.html
