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CDZSC_2015寒假新人（4）——搜索 B
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.
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)
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
思路：就是从@开始上下左右4个方向把.全部找到，dfs就好
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 char str[25][25]; 6 int vis[25][25],w,h,sum; 7 int x[]={-1,1,0,0}; 8 int y[]={0,0,-1,1}; 9 void dfs(int i,int j)10 {11 for(int k=0; k<4; k++)12 {13 if(i+x[k]<h&&i+x[k]>=0&&j+y[k]<w&&j+y[k]>=0&&vis[i+x[k]][j+y[k]]==0&&str[i+x[k]][j+y[k]]!=‘#‘)14 {15 sum++;16 vis[i+x[k]][j+y[k]]=1;17 dfs(i+x[k],j+y[k]);18 }19 }20 }21 int main()22 {23 #ifdef CDZSC_OFFLINE24 freopen("in.txt","r",stdin);25 #endif26 int i,j;27 while(scanf("%d%d",&w,&h)&&w!=0&&h!=0)28 {29 memset(vis,0,sizeof(vis));30 sum=1;31 for(i=0; i<h; i++)32 {33 scanf("%s",str[i]);34 }35 for(i=0; i<h; i++)36 {37 for(j=0; j<w; j++)38 {39 if(str[i][j]==‘@‘)40 {41 vis[i][j]=1;42 dfs(i,j);43 }44 }45 }46 printf("%d\n",sum);47 }48 return 0;49 }
CDZSC_2015寒假新人（4）——搜索 B
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