POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19254		Accepted: 10252

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) 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

一道基础的深搜题，一个人从给定的位置开始走，可以朝上下左右四个方向，只能走黑色的格子，问这个人一共可以走多少个格子

只要从起始位置开始深搜就可以了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 int w,h,start_x,start_y,ans;
 8 bool dfsed[30][30];
 9 char g[30][30];
10 
11 void dfs(int x,int y)
12 {
13     if(g[x-1][y]=='.'&&(!dfsed[x-1][y]))
14     {
15         dfsed[x-1][y]=true;
16         dfs(x-1,y);
17         ans++;
18     }
19     if(g[x+1][y]=='.'&&(!dfsed[x+1][y]))
20     {
21         dfsed[x+1][y]=true;
22         dfs(x+1,y);
23         ans++;
24     }
25     if(g[x][y-1]=='.'&&(!dfsed[x][y-1]))
26     {
27         dfsed[x][y-1]=true;
28         dfs(x,y-1);
29         ans++;
30     }
31     if(g[x][y+1]=='.'&&(!dfsed[x][y+1]))
32     {
33         dfsed[x][y+1]=true;
34         dfs(x,y+1);
35         ans++;
36     }
37 }
38 
39 int main()
40 {
41     while(scanf("%d %d",&w,&h)==2)
42     {
43         getchar();
44         if(w==0&&h==0)
45             break;
46         for(int i=1;i<=h;i++)
47             gets(&g[i][1]);
48         for(int i=0;i<=h+1;i++)
49             g[i][0]=g[i][w+1]='#';
50         for(int i=1;i<=w+1;i++)
51             g[0][i]=g[h+1][i]='#';
52         for(int i=1;i<=h;i++)
53             for(int j=1;j<=w;j++)
54                 if(g[i][j]=='@')
55                 {
56                     start_x=i;
57                     start_y=j;
58                 }
59         ans=1;
60         memset(dfsed,false,sizeof(dfsed));
61         dfs(start_x,start_y);
62         cout<<ans<<endl;
63     }
64 
65     return 0;
66 }

[C++]