HDU 1198 Farm Irrigation

这道题可能刚拿到手的话比较困难，不知如何下手。

但是分析一下貌似用搜索和并查集可以搞定。

下面提供并查集的代码，具体详解在代码注释

 

                                   Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3463    Accepted Submission(s): 1533

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
   
   
   
   

 

 

#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int M = 60;
struct node{      //保存格子水管有哪个方向的路线 
       int s;     //上 
       int x;     //下 
       int z;     //左 
       int y;     //右 
};                

node T[M][M];

int father[ 4000 ];   //虽然是二维的，但是为了方便，用一维数组来存祖先节点 
                      //( i*60 + j )i为行，j为列，得数即为这个格子的祖先的数值 
int n,m;
char s;

void initial( )
{
     for( int i=0 ; i<M ; i++ )
          for( int j=0 ; j<M ; j++ ){
               father[i*60+j] = i*60+j;     //因为输入的行列数不超过50，所以乘以了60 
               T[i][j].s = T[i][j].x = T[i][j].z = T[i][j].y = 0;
          }
}

void solve( int i , int j , char c )      //来处理每个格子的状态 
{
     if( c == 'A' ){
         T[i][j].s = 1;T[i][j].x = 0;
         T[i][j].z = 1;T[i][j].y = 0;
     }
     else if( c == 'B' ){
          T[i][j].s = 1;T[i][j].x = 0;
          T[i][j].z = 0;T[i][j].y = 1;     
     }
     else if( c == 'C' ){
          T[i][j].s = 0;T[i][j].x = 1;
          T[i][j].z = 1;T[i][j].y = 0;     
     }
     else if( c == 'D' ){
          T[i][j].s = 0;T[i][j].x = 1;
          T[i][j].z = 0;T[i][j].y = 1;     
     }
     else if( c == 'E' ){
          T[i][j].s = 1;T[i][j].x = 1;
          T[i][j].z = 0;T[i][j].y = 0;     
     }
     else if( c == 'F' ){
          T[i][j].s = 0;T[i][j].x = 0;
          T[i][j].z = 1;T[i][j].y = 1;     
     }
     else if( c == 'G' ){
          T[i][j].s = 1;T[i][j].x = 0;
          T[i][j].z = 1;T[i][j].y = 1;     
     }
     else if( c == 'H' ){
          T[i][j].s = 1;T[i][j].x = 1;
          T[i][j].z = 1;T[i][j].y = 0;     
     }else if( c == 'I' ){
          T[i][j].s = 0;T[i][j].x = 1;
          T[i][j].z = 1;T[i][j].y = 1;     
     }
     else if( c == 'J' ){
          T[i][j].s = 1;T[i][j].x = 1;
          T[i][j].z = 0;T[i][j].y = 1;     
     }
     else if( c == 'K' ){
          T[i][j].s = 1;T[i][j].x = 1;
          T[i][j].z = 1;T[i][j].y = 1;     
     }
     
}

int find( int x )
{
    return x == father[x] ? x : father[x] = find( father[x] );    
}

void merge( int a , int b )                  
{
     int x ,y;
     x = find( a );
     y = find( b );
     if( x != y ){
         father[y] = x;
     }     
}

bool fun( int i , int j )      //查看是否和4个相邻位置的格子有连通，有的话就合并 
{
     bool flag = false;
     if( T[i][j].s && T[i-1][j].x )
         {merge( (i*60)+j , (i-1)*60+j  );flag=true;}
     if( T[i][j].x && T[i+1][j].s )
         {merge( (i*60)+j , (i+1)*60+j  );flag=true;}
     if( T[i][j].z && T[i][j-1].y )
         {merge( (i*60)+j , (i*60)+j-1  );flag=true;}
     if( T[i][j].y && T[i][j+1].z )
         {merge( (i*60)+j , (i*60)+j+1  );flag=true;}
     if( flag )
         return false;
     else
         return true;
}

int main()
{
    while( scanf("%d%d",&n,&m) == 2 ){
           if( n==-1 && m==-1 )
               break;
           initial( );
           getchar( );
           for( int i=1 ; i<=n ; i++ ){
                for( int j=1 ; j<=m ; ++j ){
                     scanf("%c",&s);
                     solve( i , j , s );   
                }   
                getchar( );
           }
           int ans = 0;
           for( int i=1 ; i<=n ; i++ ){
                for( int j=1 ; j<=m ; j++ ){
                     fun(  i , j  );
                }     
           }
           for( int i=1 ; i<=n ; i++ ){
                for( int j=1 ; j<=m ; j++ ){
                     //printf("%d ",father[i*60+j]);
                     if( father[i*60+j] == i*60+j )
                         ans++;
                }     
                //printf("\n");
           }
           printf("%d\n",ans);
    }
    return 0;
}