Red and Black
Posted ⊙∽⊙Perseverance
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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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
代码:
#include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; char a[25][25]; int v[25][25]; int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0}; int m,n,e1,e2,ans; void dfs(int x,int y) { v[x][y]=1; for(int i=0;i<4;i++) {int xx,yy; xx=x+fx[i]; yy=y+fy[i]; if(xx>=0&&yy>=0&&xx<n&&yy<m&&a[xx][yy]==‘.‘&&!v[xx][yy]) //xx<n和yy<m刚开始写反了,,, {ans++; dfs(xx,yy); } } } int main() { while(scanf("%d %d",&m,&n)!=EOF&&(n||m)) {for(int i=0;i<n;i++) scanf("%s",a[i]); memset(v,0,sizeof(v)); for(int i=0;i<n;i++) {for(int j=0;j<m;j++) {if(a[i][j]==‘@‘) {e1=i; e2=j; } } } ans=1; dfs(e1,e2); printf("%d\n",ans); } return 0; }
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