Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 23194 | Accepted: 12513 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题目翻译：给出一个m*n的图，然后‘.’代表可以走的路线，‘@’代表起点，‘#’代表墙，不能走，
求：算上起点，一共有多少格子可以走？
解题思路：dfs搜索！
#include<cstdio>
int m,n,x1,y1,cot;
char G[22][22];
int mov[][2]={0, 1, 0, -1, -1, 0, 1, 0};
void dfs(int x,int y){
G[x][y]='#';
int i,X,Y;
for(i=0;i<4;i++){
X=x+mov[i][0];
Y=y+mov[i][1];
if(G[X][Y]!='#'&&X>=0&&X<n&&Y>=0&&Y<m){
cot++;
dfs(X,Y);
}
}
}
int main(){
int i,j;
while(scanf("%d%d",&m,&n),m|n){
for(i=0;i<n;++i)
scanf("%s",G[i]);
for(i=0;i<n;++i)
for(j=0;j<m;++j)
if(G[i][j]=='@'){
x1=i;y1=j;
}
cot=1;
dfs(x1,y1);
printf("%d\n",cot);
}
return 0;
}