HDU 1312 Red and Black（bfs）

Red and Black
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
 

Sample Output

45
59
6
13
题目简单翻译：
从‘@’点出发，问能到达的最多的点有多少，‘#’不可经过，计算结果包括‘@’。
 
解题思路：
广度优先搜索，直接求出到过多少个点。
 
代码：
 1 #include<cstdio>
 2 #include<cstring>
 3 
 4 using namespace std;
 5 int n,m,sx,sy;
 6 char mp[24][24];
 7 int vis[24][24];
 8 int dx[]={0,0,1,-1};
 9 int dy[]={1,-1,0,0};
10 bool check(int x,int y)
11 {
12     return x>=0&&x<n&&y>=0&&y<m;
13 }
14 struct node
15 {
16     int x,y;
17 }St[1000];
18 
19 int bfs()//手写的队列，队列的尾端就是到达的点的数量
20 {
21     memset(vis,0,sizeof vis);
22     int st=0,en=1;
23     vis[St[0].x][St[0].y]=1;
24     while(st<en)
25     {
26         node e=St[st++];
27         for(int i=0;i<4;i++)
28         {
29             node w=e;
30             w.x=e.x+dx[i],w.y=e.y+dy[i];
31             if(check(w.x,w.y)&&vis[w.x][w.y]==0&&mp[w.x][w.y]!='#')
32             {
33                 vis[w.x][w.y]=1;
34                 St[en++]=w;
35             }
36         }
37     }
38     return en;
39 }
40 
41 int main()
42 {
43     while(scanf("%d%d",&m,&n)!=EOF&&(n||m))
44     {
45         for(int i=0;i<n;i++)
46         {
47             scanf("%s",mp[i]);
48             for(int j=0;j<m;j++)
49                 if(mp[i][j]=='@')
50                     St[0].x=i,St[0].y=j;
51         }
52         printf("%d\n",bfs());
53     }
54     return 0;
55 }
Red and Black

 

posted on 2015-07-06 16:23  活力典  阅读(223)  评论(0)  编辑  收藏  举报

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