hdoj problem 2952 Counting Sheep(深搜DFS)
来源:互联网 发布:网页版淘宝 编辑:程序博客网 时间:2024/05/29 14:15
Counting Sheep
http://acm.hdu.edu.cn/showproblem.php?pid=2952
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2157 Accepted Submission(s): 1422
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.
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.
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
Notes and Constraints
0 < T <= 100
0 < H,W <= 100
Sample Input
24 4#.#..#.##.##.#.#3 5###.#..#..#.###
Sample Output
63
Source
IDI Open 2009
Recommend
gaojie | We have carefully selected several similar problems for you: 1010 1258 1312 1426 1067
/*
这是参考了人家的代码后AC的
Come On !
*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
char a[105][105];
int n,m,fx[4][4]={{1,0},{0,1},{-1,0},{0,-1}};//方向
#include<cstring>
#include<cstdlib>
using namespace std;
char a[105][105];
int n,m,fx[4][4]={{1,0},{0,1},{-1,0},{0,-1}};//方向
void DFS(int x,int y)
{
int i;
for(i=0;i<4;i++)
{
int xx=fx[i][0]+x;
int yy=fx[i][1]+y;
if(xx>=0&&yy>=0&&xx<n&&yy<m&&a[xx][yy]=='#')//递归边界
{
a[xx][yy]='.';//标记已经遍历的
DFS(xx,yy);
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int i,j,count;
scanf("%d%d",&n,&m);
getchar();
for(i=0;i<n;i++)
scanf("%s",a[i]);
count=0;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
if(a[i][j]=='#')
{
count++;
DFS(i,j);
}
printf("%d\n",count);
}
return 0;
}
{
int i;
for(i=0;i<4;i++)
{
int xx=fx[i][0]+x;
int yy=fx[i][1]+y;
if(xx>=0&&yy>=0&&xx<n&&yy<m&&a[xx][yy]=='#')//递归边界
{
a[xx][yy]='.';//标记已经遍历的
DFS(xx,yy);
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int i,j,count;
scanf("%d%d",&n,&m);
getchar();
for(i=0;i<n;i++)
scanf("%s",a[i]);
count=0;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
if(a[i][j]=='#')
{
count++;
DFS(i,j);
}
printf("%d\n",count);
}
return 0;
}
0 0
- hdoj problem 2952 Counting Sheep(深搜DFS)
- hdoj 2952 Counting Sheep 【dfs】
- HDOJ题目2952Counting Sheep (DFS)
- HDOJ 2952 Counting Sheep
- 2952 Counting Sheep【dfs】
- hdu 2952 Counting Sheep(dfs)
- HDU 2952 Counting Sheep [DFS]
- HDU-2952 Counting Sheep (DFS)
- hdu 题目2952 Counting Sheep (DFS)
- hdu 2952 Counting Sheep (dfs)
- hdu 2952 Counting Sheep(简单dfs)
- hdu 2952 Counting Sheep (DFS)
- HDU 2952 Counting Sheep (DFS)
- HDU 2952 Counting Sheep 简单dfs
- hdu 2952 Counting Sheep(dfs)
- HDU 2952 Counting Sheep ( DFS + BFS )
- HDU 2952 Counting Sheep(DFS)
- HDU2952:Counting Sheep(DFS)
- 菜鸟授徒系列之新手入门
- 遍历中序线索二叉树
- USACO 2.2.4 两只塔沃斯母牛
- liferay namespace用法
- 构造赫夫曼树
- hdoj problem 2952 Counting Sheep(深搜DFS)
- Putty and Pycharm的风格configuration
- UVALive-6665-Dragons Cruller(Dij+Hash)
- 树莓派&nodejs相关开发过程
- Ubuntu-ADT-连接Anroid真机-无法识别问题
- ubuntu修改、删除用户
- Raspberry Pi学习笔记
- USACO 2.2.5 挤牛奶
- hdoj 1181 problem变形课(并查集)