POJ1979 Red and Black

速刷一道DFS

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.

@是起点，#不能走，'.'是正常道路，问从起点能到达多少格子

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Source

Japan 2004 Domestic

刷道普通的DFS，复习模板

 /*by SilverN*/
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 const int mxn=;
 char mp[mxn][mxn];
 int ans;
 int W,H;
 void dfs(int x,int y){
     if(x< || x>H || y< || y>W)return;
     if(mp[x][y]=='#')return;
     ans++;
     mp[x][y]='#';
     dfs(x+,y);
     dfs(x,y+);
     dfs(x-,y);
     dfs(x,y-);
     return;
 }
 int main(){
     while(scanf("%d%d",&W,&H) && W && H){
         memset(mp,' ',sizeof(mp));
         ans=;
         int i,j;
         for(i=;i<=H;i++){
             scanf("%s",mp[i]+);
         }
         int sx,sy;
         bool flag=;
         for(i=;i<=H;i++){
               for(j=;j<=W;j++)
                   if(mp[i][j]=='@'){
                       sx=i;sy=j;
                       flag=;
                       break;
                   }
             if(flag)break;
         }
         dfs(sx,sy);
         printf("%d\n",ans);
     }
     return ;
 }