 Counting Sheep

Problem Description
A while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. Then one day my grandmother suggested I tried counting sheep after I'd gone to bed. As always when my grandmother suggests things, I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed.Creative as I am, that wasn't going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps A and C are in the same flock.
Now, I've got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I've decided I need another computer program that does the counting for me. Then I'll be able to just start both these programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.
Input
The first line of input contains a single number T, the number of test cases to follow.Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.
Output
For each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the problem description.Notes and Constraints
0 < T <= 100
0 < H,W <= 100Sample Input
2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###Sample Output
6
3
用dfs就应该可以了。
#include<bits/stdc++.h>
using namespace std;
int n,m,ans,t;
char c[101][101];
bool f[101][101];
int dx[5]={0,1,0,1,0},dy[5]={0,0,1,0,1};
void dfs(int x,int y){
int xx,yy,k;
for(k=1;k<=4;k++){
xx=x+dx[k];
yy=y+dy[k];
if(xx<1xx>nyy<1yy>mf[xx][yy]c[xx][yy]=='.') continue;
f[xx][yy]=1;
dfs(xx,yy);
}
}
int main(){
int i,j;
scanf("%d",&t);
while(t){
ans=0;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++) scanf("%s",c[i]+1);
for(i=1;i<=n;i++) for(j=1;j<=m;j++) f[i][j]=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(c[i][j]=='#'&&!f[i][j]){
ans++;
dfs(i,j);
}
printf("%d\n",ans);
}
return 0;
}
如果出错了，请告诉我