Farm Irrigation
Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.
Figure 1
Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map
ADC
FJK
IHE
then the water pipes are distributed like
Figure 2
Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.
Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?
Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.
Input
There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.
Output
For each test case, output in one line the least number of wellsprings needed.
Sample Input
2 2
DK
HF
3 3
ADC
FJK
IHE
-1 -1
Sample Output
2
3
题目类型:DFS/BFS
算法分析:统计程序中调用的DFS/BFS的次数即可,搜索中判断转移的过程是当前所在位置存在向周围的4个方向移动的水管,然后再判断移动后的位置的水管是否有与移动前的位置的水管相反的(既保证相邻草坪存在水管路径)。注意可以使用%运算来简化代码
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#include <iostream> #include <fstream> #include <cstring> using namespace std; const int maxn = 100 + 36; const int dx[] = {-1, 0, 1, 0}; const int dy[] = {0, 1, 0, -1}; const bool flag[26][6] = { {1, 0, 0, 1} , {1, 1, 0, 0} , {0, 0, 1, 1} , {0, 1, 1, 0} , {1, 0, 1, 0} , {0, 1, 0, 1} , {1, 1, 0, 1} , {1, 0, 1, 1} , {0, 1, 1, 1} , {1, 1, 1, 0} , {1, 1, 1, 1}}; char ans[maxn][maxn]; bool visited[maxn][maxn]; int row, col; inline bool OkMove (int x, int y, int val) { if (x >= 0 && x <= row - 1 && y >= 0 && y <= col - 1 && !visited[x][y] && flag[ans[x-dx[val]][y-dy[val]]-'A'][val] && flag[ans[x][y]-'A'][(val+2)%4]) return true; return false; } void DFS (int x, int y) { int i; for (i = 0; i < 4; i++) { if (OkMove (x + dx[i], y + dy[i], i)) { int val_x = x + dx[i], val_y = y + dy[i]; visited[val_x][val_y] = true; DFS (val_x, val_y); } } } int main() { // ifstream cin ("aaa.txt"); while (cin >> row >> col) { if (row < 0 || col < 0) break; int i; for (i = 0; i < row; i++) cin >> ans[i]; int sum = 0, j; memset (visited, false, sizeof (visited)); for (i = 0; i < row; i++) for (j = 0; j < col; j++) if (!visited[i][j]) { visited[i][j] = true; DFS (i, j); sum++; } cout << sum << endl; } return 0; } |
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