Farm Irrigation HDU

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10286    Accepted Submission(s): 4523

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 2DKHF3 3ADCFJKIHE-1 -1

 

Sample Output

23

题目分析：因为题目是要求有几个独立的分开块，很明显的并查集特性。两重循环扫一遍，如果两个块之间能连通就加入到一个集合当中。最后看有多少个独立集合就行了

代码：

#include <iostream>  #include <stdio.h>  #include <string>  using namespace std;    const int M=55;  char ti[M][M];  int uf[M*M];  int isc[M*M];    int shang[]={1,1,0,0,1,0,1,1,0,1,1};  int xia[]={0,0,1,1,1,0,0,1,1,1,1};  int zuo[]={1,0,1,0,0,1,1,1,1,0,1};  int you[]={0,1,0,1,0,1,1,0,1,1,1};    int find(int i)  {      if(i==uf[i])          return i;      return uf[i]=find(uf[i]);  }    void merge(int i,int j)  {      int t1=find(i);      int t2=find(j);      if(t1==t2)          return;      uf[t2]=t1;  }    int main()  {      int n,m;      while(1)      {          scanf("%d%d",&n,&m);          if(n<0&&m<0)              break;          for(int i=1;i<=n*m;i++)          {              uf[i]=i;              isc[i]=0;          }          for(int i=1;i<=n;i++)              scanf("%s",ti[i]+1);          for(int i=1;i<=n;i++)          {              for(int j=1;j<m;j++)              {                  if(you[ti[i][j]-'A']&&zuo[ti[i][j+1]-'A'])                      merge((i-1)*m+j,(i-1)*m+j+1);              }          }          for(int i=1;i<=m;i++)          {              for(int j=1;j<n;j++)              {                  if(xia[ti[j][i]-'A']&&shang[ti[j+1][i]-'A'])                      merge((j-1)*m+i,j*m+i);              }          }          int co=0;          for(int i=1;i<=n*m;i++)                {                                                if(!isc[find(i)])                           {                  co++;                  isc[find(i)]=1;              }          }          printf("%d\n",co);      }      return 0;  }