Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 17061 | Accepted: 8996 |
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
#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;
char s[25][25];
bool visit[25][25];
int col,row;
int ans;
void dfs(int x,int y)
{
visit[x][y]=1;
if(x>0&&!visit[x-1][y]&&s[x-1][y]=='.')
{
ans++;
dfs(x-1,y);
}
if(y>0&&!visit[x][y-1]&&s[x][y-1]=='.')
{
ans++;
dfs(x,y-1);
}
if(x+1<row&&!visit[x+1][y]&&s[x+1][y]=='.')
{
ans++;
dfs(x+1,y);
}
if(y+1<col&&!visit[x][y+1]&&s[x][y+1]=='.')
{
ans++;
dfs(x,y+1);
}
}
int main()
{
int i,j,k,T;
int p,q;
while(cin>>col>>row,row||col)
{
ans = 1;
memset(s,0,sizeof(s));
memset(visit,0,sizeof(visit));
for(i=0;i<row;i++)
{
//getchar();
for(j=0;j<col;j++)
{
cin>>s[i][j];
if(s[i][j]=='@')
{
p = i;
q = j;
}
}
}
dfs(p,q);
cout<<ans<<endl;
}
system("pause");
return 0;
}
//小花熊的
#include<stdio.h>
char m[20][20];
int r,c,count;
void cnt(int i,int j)//统计连续黑砖的块数
{
if(m[i][j]=='#'||(i<0||j<0)||(i>r-1||j>c-1))//边界条件，除去
return;
m[i][j]='#';//发现了一个新的黑砖，置'#'，下次不在访问
count++; //count+1
cnt(i,j-1);//往左寻找
cnt(i-1,j);//往上寻找
cnt(i,j+1);//往右寻找
cnt(i+1,j);//往下寻找
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&c,&r),r||c){
for(count=i=0;i<r;++i){
getchar();
for(j=0;j<c;++j){
m[i][j]=getchar();
if(m[i][j]=='@'){//查找起始点
x=i;
y=j;
}
}
}
cnt(x,y);
printf("%d\n",count);
}
return 0;
}