poj 1979 Red and Black（dfs）

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) 
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

题意：找出@能够到达的所有点的个数
思路：dfs和递归都能做
dfs代码：
 1 #include <iostream>
 2 #include<cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 using namespace std;
 7 char s[25][25];
 8 int a[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
 9 int ans,n,m;
10 int dfs(int x,int y)
11 {
12     for(int i=0;i<4;i++)
13     {
14         int tx=x+a[i][0];
15         int ty=y+a[i][1];
16         if(tx>=m||tx<0||ty<0||ty>=n)
17             continue;
18         if(s[tx][ty]=='.')
19         {
20             s[tx][ty]='#';
21             dfs(tx,ty);
22             ans++;
23         }
24     }
25     return 0;
26 }
27 int main()
28 {
29     while(~scanf("%d%d",&n,&m))
30     {
31         if(n==0||m==0)
32             break;
33         int flag=1;
34         int tx,ty;
35         for(int i=0;i<m;i++)
36         {
37             scanf("%s",s[i]);
38             if(flag)
39             {
40                 for(int j=0;j<n;j++)
41                 {
42                     if(s[i][j]=='@')
43                     {
44                         tx=i,ty=j;
45                         flag=0;
46                         break;
47                     }
48                 }
49             }
50         }
51         ans= 1;
52         dfs(tx,ty);
53         printf("%d\n",ans);
54     }
55     return 0;
56 }
View Code

递归代码：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <stdio.h>
 5 int n,m;
 6 char data[30][30];
 7 using namespace std;
 8 int f(int i,int j)
 9 {
10     if(i<0||j>n-1||i>m-1||j<0)
11     return 0;
12     if(data[i][j]=='#')
13     return 0;
14     data[i][j]='#';
15     return 1+f(i-1,j)+f(i+1,j)+f(i,j-1)+f(i,j+1);
16 }
17 int main()
18 {
19     int x,y;
20     while(~scanf("%d%d",&n,&m)&&(n!=0||m!=0))
21     {
22         getchar();
23         for(int i=0;i<m;i++)
24         scanf("%s",data[i]);
25         for(int i=0;i<m;i++)
26         for(int j=0;j<n;j++)
27         if(data[i][j]=='@')
28         {
29             x=i;y=j;
30         }
31         cout<<f(x,y)<<endl;
32     }
33     return 0;
34 }
View Code

 

posted @ 2016-12-22 20:41  Wally的博客  阅读(112)  评论(0)  编辑  收藏  举报