HDU - 1312 ： Red and Black

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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.
InputThe 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)
OutputFor 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

@可以通过上下左右通到的.有几个。可以DFS也可以BFS，只需把图遍历即可。

 1 #include<iostream>
 2 #include<string.h>
 3 
 4 using namespace std;
 5 
 6 char a[22][22];
 7 int s;
 8 
 9 void DFS(int x1,int y1)
10 {
11     if(a[x1][y1]==‘.‘||a[x1][y1]==‘@‘)
12     {
13         a[x1][y1]=‘#‘;
14         s++;
15     DFS(x1+1,y1);
16     DFS(x1-1,y1);
17     DFS(x1,y1+1);
18     DFS(x1,y1-1);
19     return;
20     }
21 }
22 
23 int main()
24 {
25     int x,y;
26     while(scanf("%d %d",&y,&x),x||y)
27     {
28         int x1,y1;
29         s=0;
30         memset(a,0,sizeof(a));
31         for(int i=1;i<=x;i++)
32         {
33             getchar();
34             for(int j=1;j<=y;j++)
35             {
36                 scanf("%c",&a[i][j]);
37                 //cout<<i<<" "<<j<<endl;
38                 if(a[i][j]==‘@‘)
39                 {
40                     x1=i;
41                     y1=j;
42                 }
43             }
44         }
45         DFS(x1,y1);
46         cout<<s<<endl;
47     }
48     
49     return 0;
50 }

HDU - 1312 ： Red and Black

标签：turn   one   name   data   memset   exactly   sample   ons   eof   

原文地址：http://www.cnblogs.com/xibeiw/p/7265271.html