Red and Black
Red and Black |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 118 Accepted Submission(s): 92 |
Problem 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.
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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.
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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).
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Sample Input 6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
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Sample Output 45 59 6 13
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Source Asia 2004, Ehime (Japan), Japan Domestic
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Recommend Eddy |
#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
char mp[21][21];
int dis[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int n,m;
int ans;
void dfs(int x,int y)
{
ans++;
mp[x][y]='#';
for(int i=0; i<4; i++)
{
int tx=x+dis[i][0];
int ty=y+dis[i][1];
if(tx>=0&&tx<m&&ty>=0&&ty<n&&mp[tx][ty]=='.')
{
dfs(tx,ty);
}
}
return;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
return 0;
int x,y;
for(int i=0; i<m; i++)
scanf("%s",mp[i]);
for(int i=0; i<m; i++)
{
for(int j=0; j<n; j++)
{
if(mp[i][j]=='@')
{
x=i;
y=j;
}
}
}
ans=0;
dfs(x,y);
printf("%d\n",ans);
}
return 0;
}