hdoj_2952Counting Sheep
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)
Keep it simple!