HDU 1312 Red and Black (DFS)

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

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

7 7

..#.#.. ..#.#..

###.###

...@...

###.###

..#.#..

0 0

Sample Output

45

59

6

13

要小心它的输入

行列是相反的

坑在这里找了好多次TAT

 1 #include<stdio.h>
 2 #include<iostream>
 3 using namespace std;
 4 char map[25][25];
 5 int n,m;
 6 int num;
 7 int rx[]={0,0,1,-1};
 8 int ry[]={1,-1,0,0};
 9 void dfs(int i,int j)
10 {
11     map[i][j]='#';
12     num++;
13     int t,x,y;
14     for(t=0;t<4;t++)
15     {
16         x=i+rx[t];
17         y=j+ry[t];
18         if(x>=0 && x<n && y>=0 && y<m && map[x][y]=='.')
19             dfs(x,y);    
20     }
21 }
22 int main()
23 {
24     int i,j;
25     int si,sj;
26     while(scanf("%d%d",&m,&n)!=EOF)//原来是行数和列数反了...我真是醉了....
27     {
28         
29         if(m==0&&n==0)break;
30         for(i=0;i<n;i++)
31             for(j=0;j<m;j++)
32             {
33                 cin>>map[i][j];
34                 if(map[i][j]=='@')
35                 {
36                     si=i;
37                     sj=j;
38                 }
39             }
40         num=0;
41         dfs(si,sj);
42         printf("%d
",num);
43     }
44     return 0;
45 }

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原文地址：https://www.cnblogs.com/Annetree/p/5641818.html