hdu 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10133 Accepted Submission(s): 6321

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

....#.

.....#

......

#@...#

.#..#.

0 0

Sample Output

Source

Asia 2004, Ehime (Japan), Japan Domestic

PS:简单题

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<queue>
 7 using namespace std;
 8 int n,m;
 9 int xi,yj;
10 char map[25][25];
11 int vis[25][25];
12 int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//上下左右
13 struct node{
14     int x;
15     int y;
16 };
17 
18 bool check(int x,int y)
19 {
20     if(x>=1&&x<=n && y>=1&&y<=m && map[x][y]=='.' && !vis[x][y])
21     return true;
22     return false;
23 }
24 
25 int bfs()
26 {
27     memset(vis,0,sizeof(vis));
28     int i,sum=1;
29     node p,q;
30     queue<struct node>Q;
31     p.x=xi;
32     p.y=yj;
33     Q.push(p);
34     while(!Q.empty())
35     {
36         p=Q.front();
37         Q.pop();
38 
39         for(i=0;i<4;i++)
40         {
41             q.x=p.x+dir[i][0];
42             q.y=p.y+dir[i][1];
43             if(check(q.x,q.y))
44             {
45                 vis[q.x][q.y]=1;
46                 sum++;
47                 Q.push(q);
48             }
49         }
50     }
51     return sum;
52 }
53 
54 
55 int main()
56 {
57     int i,j;
58     int count;
59     //freopen("in.txt","r",stdin);
60     while(~scanf("%d%d",&m,&n))
61     {
62 
63         if(!n&&!m)
64         break;
65         for(i=1;i<=n;i++)
66         {
67             getchar();
68             for(j=1;j<=m;j++)
69             {
70                 scanf("%c",&map[i][j]);
71                 if(map[i][j]=='@')
72                 {
73                     xi=i;
74                     yj=j;
75                 }
76             }
77         }
78         count=bfs();
79         printf("%d
",count);
80     }
81     return 0;
82 }

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