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2014
02-24

HDU 2952-Counting Sheep -DFS-[解题报告]HOJ

Counting Sheep

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

6
3

这个题目也可以使用广度搜索方式解题。这里写出深度方法的代码和思路

#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;
}

解题参考:http://blog.csdn.net/mxway/article/details/8581620


,
  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。