Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 19263 | Accepted: 10259 |
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)
The end of the input is indicated by a line consisting of two zeros.
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
这是一道简单的dfs，我手残在忘记把文件输入的语句 给注释了，结果就跪了。
下面是代码：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
int dir[8][2]={{-1,0},{0,1},{1,0},{0,-1}};
int w,h,ans;
char map[30][30];
void dfs(int x,int y){
if(x<0||x>=h||y<0||y>=w)
return;
for(int i=0;i!=4;++i){
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(map[xx][yy]=='.'){
map[xx][yy]='1';
dfs(xx,yy);
}
}
return;
}
int main(){
//freopen("input.txt","r",stdin);
// freopen("output.txt","w",stdout);
int sx,sy;
while(scanf("%d%d",&w,&h)!=EOF&&w&&h){
for(int i=0;i!=h;++i){
scanf("%s",map[i]);
for(int j=0;j!=w;++j){
if(map[i][j]=='@'){
sx=i;
sy=j;
}
}
}
map[sx][sy]='1';
ans=0;
dfs(sx,sy);
for(int i=0;i!=h;++i)
for(int j=0;j!=w;++j){
if(map[i][j]=='1')
++ans;
}
cout<<ans<<endl;
}
}