Counting Sheep
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1281 Accepted Submission(s): 836
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
2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.###
#include <iostream>
#include <cstring>
using namespace std;
#pragma warning(disable : 4996)
#define MAX 105
char maps[MAX][MAX];
int h, w;
const int moves[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
void dfs(int x, int y)
{
maps[x][y] = '.';
for(int i = 0; i < 4; i++)
{
int p = x + moves[i][0];
int q = y + moves[i][1];
if(p >= 1 && p <= h && q >= 1 && q <= w && maps[p][q] == '#')
{
dfs(p, q);
}
}
}
int main()
{
freopen("in.txt", "r", stdin);
int t,cnt;
cin >> t;
while(t--)
{
cin >> h >> w;
for(int i = 1; i <= h; i++)
{
for(int j = 1; j <= w; j++)
{
cin >> maps[i][j];
}
}
cnt = 0;
for(int i = 1; i <= h; i++)
{
for(int j = 1; j <= w; j++)
{
if(maps[i][j] == '#')
{
dfs(i, j);
cnt++;
}
}
}
cout << cnt << endl;
}
}最新版VS2012，使用freopen显示error，用下面的屏蔽
#pragma warning(disable : 4996)