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 <= 100

Sample Input
2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###

Sample Output
6
3

2个回答

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<1||xx>n|yy<1||yy>m||f[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

shunfurh
shunfurh 谢谢,但是你的程序不行。
一年多之前 回复
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