HDU1198 Farm Irrigation

Farm Irrigation


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7740    Accepted Submission(s): 3314




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 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

             题目大意：  主要是说，蓝色的是水管，绿色的是草坪，

          有11块不同类型的草坪。然后，输入各块草坪的字母后，

                输出有几块水管连通的草坪！

#include<stdio.h>char b[11][5]={"1010","1001","0110","0101","1100",        "0011","1011","1110","0111","1101","1111"};// 对每块草坪进行描述，上下左右各个方向有水管为1，否则为0； char a[55][55];int ff[55][55],m,n;int find(int x)  //对草坪进行查找父节点，对线路进行压缩！ {if(x!=ff[x/n][x%n]) ff[x/n][x%n]=find(ff[x/n][x%n]); return ff[x/n][x%n];}void union1(int x,int y)//对两个点进行联合！ {x=find(x);y=find(y);if(x!=y) ff[y/n][y%n]=x;}void judge(int i,int j)//判断两块草坪是不是连通的！ {if(j>0&&b[a[i][j]-'A'][2]=='1'&&b[a[i][j-1]-'A'][3]=='1')union1(i*n+j,i*n+(j-1));if(i>0&&b[a[i][j]-'A'][0]=='1'&&b[a[i-1][j]-'A'][1]=='1')union1(i*n+j,(i-1)*n+j);}int main(){int i,j,k,l;while(scanf("%d%d",&m,&n)!=EOF&&(m!=-1||n!=-1)){for(i=0;i<m;i++){scanf("%s",a[i]);    for(j=0;j<n;j++) ff[i][j]=i*n+j;}for(i=0;i<m;i++)  for(j=0;j<n;j++)    judge(i,j);    k=0;    for(i=0;i<m;i++)      for(j=0;j<n;j++)      {      if(ff[i][j]==i*n+j)      k++;   //寻找父节点，有几个父节点就有几块连通图！   }  printf("%d\n",k);}return 0; }