Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12935    Accepted Submission(s): 8006

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Asia 2004, Ehime (Japan), Japan Domestic

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Eddy   |   We have carefully selected several similar problems for you:  1241 1016 1010 1372 1242

这是我写搜索的第一题，dfs bfs都是在这一题上学会的。从今天开始我要开始搞搜索了，我必须全面提高自己的能力，所有的类型都得有个差不多，不然根本没法和他们讨论。

bfs & dfs

 1 #include<math.h>
 2 #include<stdio.h>
 3 #include<queue>
 4 #include<string.h>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define N 1234
 9 struct point
10 {
11     int x,y;
12 }st;
13
14 int n,m,ans;
15 int dx[]={0,0,1,-1};
16 int dy[]={1,-1,0,0};
17 char mat[N][N];
18 int  vis[N][N];
19
20 void bfs()
21 {
22     queue<point>q;
23     q.push(st);vis[st.x][st.y]=1;
24     while(!q.empty())
25     {
26         point cur=q.front();
27         q.pop();
28         ans++;
29         for(int i=0;i<4;i++)
30         {
31             point next=cur;
32             next.x+=dx[i];next.y+=dy[i];
33             if(next.x<1||next.x>n||next.y<1||next.y>m)continue;
34             if(mat[next.x][next.y]==‘#‘||vis[next.x][next.y]==1)continue;
35             q.push(next);vis[next.x][next.y]=1;
36         }
37     }
38 }
39 void dfs(int x,int y)
40 {
41     if(x<1||x>n||y<1||y>m)return;
42     if(mat[x][y]==‘#‘||vis[x][y]==1)return;
43     vis[x][y]=1;
44     ans++;
45     for(int i=0;i<4;i++)
46         dfs(x+dx[i],y+dy[i]);
47 }
48
49 int main()
50 {
51     while(~scanf("%d%d",&m,&n)&&(m+n))
52     {
53         memset(vis,0,sizeof(vis));
54         for(int i=1;i<=n;i++)
55             for(int j=1;j<=m;j++)
56             {
57                 scanf(" %c",&mat[i][j]);
58                 if(mat[i][j]==‘@‘)
59                 {
60                     st.x=i;st.y=j;
61                 }
62             }
63             ans=0;
64             bfs();
65 //            dfs(st.x,st.y);
66             cout<<ans<<endl;
67     }
68     return 0;
69 }