HDu 1198 Farm Irrigation

HDu 1198 Farm Irrigation

 

Farm Irrigation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 7427 Accepted Submission (s): 3185


Problem Description 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 could wellsprings shocould 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 2DKHF3 3ADCFJKIHE-1 -1

Sample Output

23

 

I would like to tell several fields that irrigation can be performed if there is a pipe connection.

The main analysis is to find the UNICOM block, a Unicom block, and an irrigation device. You just need a few Unicom blocks.

 

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# Include using namespace std; char s [1009] [1009]; int vis [1009] [1009]; int m, n; int dis [4] [2] = {, 0,-, 0,-}; // int up [] = {, 0,, 1}; int dw [] = {, 1}; int le [] =,, 1}; int ri [] = {, 1}; void dfs (int x, int y) {if (vis [x] [y]) return; vis [x] [y] = 1; for (int I = 0; I <4; I ++) {int xx = x + dis [I] [0]; int yy = y + dis [I] [1]; if (xx <0 | xx> = m | yy <0 | yy> = N) continue; if (I = 0 & ri [s [x] [y]-'a'] & le [s [xx] [yy]-'a']) dfs (xx, yy ); if (I = 1 & le [s [x] [y]-'a'] & ri [s [xx] [yy]-'a']) dfs (xx, yy ); if (I = 2 & dw [s [x] [y]-'a'] & up [s [xx] [yy]-'a']) dfs (xx, yy ); if (I = 3 & up [s [x] [y]-'a'] & dw [s [xx] [yy]-'a']) dfs (xx, yy) ;}} int main () {while (~ Scanf (% d, & m, & n) {if (n <0 | m <0) break; memset (vis, 0, sizeof vis ); for (int I = 0; I