2014
02-24

Counting Sheep

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.

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.

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.

2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###

6
3

#include<stdio.h>
#include<string.h>

int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};//四个方向搜索
char map[110][110];
int flag[110][110];//flag[x][y]>0表示(x,y)处的羊已划分到某个区域中。
int H,W;

int check(int x,int y){
if(x>=0 && y>=0 && x<H && y<W)return 1;
return 0;
}

void dfs(int x,int y,int cnt){
int i,dx,dy;
for(i=0;i<4;i++){
dx=x+dir[i][0];
dy=y+dir[i][1];
//从(x,y)点搜索到(dx,dy)，并且(dx,dy)处的羊还未划分到某个区域中。
if(check(dx,dy) && flag[dx][dy]==0 && map[dx][dy]=='#'){
flag[dx][dy]=cnt;//标记已划分
//从(dx,dy)开始搜索查看(dx,dy)的"四周"是否有未标记的点
dfs(dx,dy,cnt);
}
}
}

int main(){
int T,i,j,cnt;
scanf("%d",&T);
while(T--){
memset(flag,0,sizeof(flag));
scanf("%d%d",&H,&W);
for(i=0;i<H;i++)scanf("%s",map[i]);
cnt=0;
for(i=0;i<H;i++){
for(j=0;j<W;j++){
//(i,j)处有羊，并且尚未标记。一个新的区域。
if(map[i][j]=='#' && flag[i][j]==0){
cnt++;//区域数加1
flag[i][j]=cnt;
dfs(i,j,cnt);
}
}
}
printf("%d\n",cnt);
}
return 0;
}

1. L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-1]）这个地方也也有笔误
应改为L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-2]）

2. 5.1处，反了；“上一个操作符的优先级比操作符ch的优先级大，或栈是空的就入栈。”如代码所述，应为“上一个操作符的优先级比操作符ch的优先级小，或栈是空的就入栈。”

3. 很高兴你会喜欢这个网站。目前还没有一个开发团队，网站是我一个人在维护，都是用的开源系统，也没有太多需要开发的部分，主要是内容整理。非常感谢你的关注。