HDU1312-Red and Black-DFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19690    Accepted Submission(s): 11965

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

Sample Output

45

59

6

13

代码:

 #include<bits/stdc++.h>
 using namespace std;
 int direct [][]={-,,,,,,,-};
 char str[][];
 bool flag[][];
 int w,h,ans;
 void DFS(int x,int y){
     for(int i=;i<;i++){
         int p=x+direct[i][];
         int q=y+direct[i][];
         if(p>=&&p<h&&q>=&&q<w&&flag[p][q]==&&str[p][q]=='.'){
             ans++;
             flag[p][q]=;
             DFS(p,q);
         }
     }
 }
 int main(){
     int Dx,Dy;
     while(~scanf("%d%d",&w,&h)){
         if(w==&&h==)break;
         memset(flag,,sizeof(flag));
         getchar();
         for(int i=;i<h;i++){
             for(int j=;j<w;j++){
                 scanf("%c",&str[i][j]);
                 if(str[i][j]=='@'){
                     Dx=i;
                     Dy=j;
                 }
             }getchar();
         }
         ans=;
         flag[Dx][Dy]=;
         DFS(Dx,Dy);
         printf("%d\n",ans);
     }
     return ;
 }