dfs水题_hdu1312，人间不值得

Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28227    Accepted Submission(s): 17064   Problem Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. Write a program to count the number of black tiles which he can reach by repeating the moves described above.   Input The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. '.' - a black tile '#' - a red tile '@' - a man on a black tile(appears exactly once in a data set)   Output For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).   Sample Input 6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0   Sample Output 45 59 6 13 离变成富婆又近了一步的AC代码： #include #include #include using namespace std; int h,w; vectorpic; int idx[22][22]; int key=0; void dfs(int r,int c) {     if(r<0||r>=h||c<0||c>=w)         return;     if(pic[r][c]=='#'||idx[r][c]!=0)         return ;     idx[r][c]=1;     key++;     for(int pr=-1;pr<=1;pr++)         for(int pc=-1;pc<=1;pc++){             if((pc&&!pr)||(!pc&&pr))   //这里应该注意搜索的方向，是十字                 dfs(r+pr,c+pc);         } } int main() {     string word;     while((cin>>w>>h)&&w&&h){//w表示列,h表示行          key=0;         pic.clear();         for(int i=0;i             cin>>word;             pic.push_back(word);             fill(idx[i],idx[i]+w,0);         }         for(int i=0;i             for(int j=0;j                 if(idx[i][j]==0&&pic[i][j]=='@'){                     dfs(i,j);                 }                                  }             cout<     }     return 0; }