zoj 2412 Farm Irrigation

ZOJ Problem Set - 2412

Farm Irrigation

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

Author: ZHENG, Lu

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Source: Zhejiang University Local Contest 2005
Submit    Status

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// 1863626 2009-05-10 21:31:51 Wrong Answer  2412 C++ 0 200 Wpl 
// 1864787 2009-05-11 18:23:54 Accepted  2412 C++ 10 200 Wpl 
#include  < iostream >
#define  MAX 55
using   namespace  std;
int  n,m,un[MAX * MAX];
char  map[MAX][MAX];
int  dir[ 2 ][ 2 ] = {{ 0 , - 1 },{ - 1 , 0 }};
void  Init()
{
     int  i,j;
     for (i = 0 ;i < m;i ++ )
         for (j = 0 ;j < n;j ++ )
        {
            un[i * n + j] = i * n + j;
            cin >> map[i][j];
        }
}
bool  Bound( int  x, int  y)
{
     if (x >= 0 && y >= 0 && x < m && y < n)
         return   true ;
     else
         return   false ;
}
int  Find( int  x)
{
     return  un[x];
}
void  Merge( int  a, int  b)
{
     int  mmax,mmin,i,temp;
    mmax = Find(a);
    mmin = Find(b);
     if (mmax == mmin)
         return  ;
     if (mmin > mmax)
    {
        temp = mmin;
        mmin = mmax;
        mmax = temp;
    }
     for (i = 0 ;i < n * m;i ++ )
    {
         if (un[i] == mmax)
            un[i] = mmin;
    }
}
int  main()
{
     int  i,j,k;
     while (scanf( " %d%d " , & m, & n) != EOF)
    {
         if (m < 0 || n < 0 )
             break ;
        Init();
         int  ti,tj,a,b;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 for (k = 0 ;k < 2 ;k ++ )
                {
                    ti = i + dir[k][ 0 ];
                    tj = j + dir[k][ 1 ];
                     if ( ! Bound(ti,tj))
                         continue ;
                    a = i * n + j;   // 我用行存储
                    b = ti * n + tj;
                     if (k == 0 )   // 向左
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' C ' || map[i][j] == ' F ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' I ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' B ' || map[ti][tj] == ' D ' || map[ti][tj] == ' F ' || map[ti][tj] == ' G ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                     else       // 向上
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' B ' || map[i][j] == ' E ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' J ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' C ' || map[ti][tj] == ' D ' || map[ti][tj] == ' E ' || map[ti][tj] == ' H ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                }
            }
         int  sum = 0 ;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 if (un[i * n + j] == (i * n + j))
                    sum ++ ;
            }
        printf( " %d\n " ,sum);
    }
     return   0 ;
}

好好反思这个代码为什么错误??????

 

Problem :  1198  ( Farm Irrigation )     Judge Status : Wrong Answer
RunId :  1357129     Language : C ++     Author : forever4444
Code Render Status : Rendered By HDOJ C ++  Code Rander Version  0.01  Beta
// 1863626 2009-05-10 21:31:51 Wrong Answer  2412 C++ 0 200 Wpl 
#include  < iostream >
#define  MAX 55
using   namespace  std;
int  n,m,un[MAX * MAX];
char  map[MAX][MAX];
int  dir[ 2 ][ 2 ] = {{ 0 , - 1 },{ - 1 , 0 }};
void  Init()
{
     int  i,j;
     for (i = 0 ;i < m;i ++ )
         for (j = 0 ;j < n;j ++ )
        {
            un[i * m + j] = i * m + j;
            cin >> map[i][j];
        }
}
bool  Bound( int  x, int  y)
{
     if (x >= 0 && y >= 0 && x < m && y < n)
         return   true ;
     else
         return   false ;
}
int  Find( int  x)
{
     return  un[x];
}
void  Merge( int  a, int  b)
{
     int  mmax,mmin,i,temp;
    mmax = Find(a);
    mmin = Find(b);
     if (mmax == mmin)
         return  ;
     if (mmin > mmax)
    {
        temp = mmin;
        mmin = mmax;
        mmax = temp;
    }
     for (i = 0 ;i < n * m;i ++ )
    {
         if (un[i] == mmax)
            un[i] = mmin;
    }
}
int  main()
{
     int  i,j,k;
     while (scanf( " %d%d " , & m, & n) != EOF)
    {
         if (m < 0 || n < 0 )
             break ;
        Init();
         int  ti,tj,a,b;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 for (k = 0 ;k < 2 ;k ++ )
                {
                    ti = i + dir[k][ 0 ];
                    tj = j + dir[k][ 1 ];
                    a = i * m + j;   // 我用行存储
                    b = ti * m + tj;
                     if ( ! Bound(ti,tj))
                         continue ;
                     if (k == 0 )   // 向左
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' C ' || map[i][j] == ' F ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' I ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' B ' || map[ti][tj] == ' D ' || map[ti][tj] == ' F ' || map[ti][tj] == ' G ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                     else       // 向上
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' B ' || map[i][j] == ' E ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' J ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' C ' || map[ti][tj] == ' D ' || map[ti][tj] == ' E ' || map[ti][tj] == ' H ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                }
            }
         int  sum = 0 ;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 if (un[i * m + j] == (i * m + j))
                    sum ++ ;
            }
        printf( " %d\n " ,sum);
    }
     return   0 ;
}
// 1863626 2009-05-10 21:31:51 Wrong Answer  2412 C++ 0 200 Wpl 
#include  < iostream >
#define  MAX 55
using   namespace  std;
int  n,m,un[MAX * MAX];
char  map[MAX][MAX];
int  dir[ 2 ][ 2 ] = {{ 0 , - 1 },{ - 1 , 0 }};
void  Init()
{
     int  i,j;
     for (i = 0 ;i < m;i ++ )
         for (j = 0 ;j < n;j ++ )
        {
            un[i * m + j] = i * m + j;
            cin >> map[i][j];
        }
}
bool  Bound( int  x, int  y)
{
     if (x >= 0 && y >= 0 && x < m && y < n)
         return   true ;
     else
         return   false ;
}
int  Find( int  x)
{
     return  un[x];
}
void  Merge( int  a, int  b)
{
     int  mmax,mmin,i,temp;
    mmax = Find(a);
    mmin = Find(b);
     if (mmax == mmin)
         return  ;
     if (mmin > mmax)
    {
        temp = mmin;
        mmin = mmax;
        mmax = temp;
    }
     for (i = 0 ;i < n * m;i ++ )
    {
         if (un[i] == mmax)
            un[i] = mmin;
    }
}
int  main()
{
     int  i,j,k;
     while (scanf( " %d%d " , & m, & n) != EOF)
    {
         if (m < 0 || n < 0 )
             break ;
        Init();
         int  ti,tj,a,b;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 for (k = 0 ;k < 2 ;k ++ )
                {
                    ti = i + dir[k][ 0 ];
                    tj = j + dir[k][ 1 ];
                    a = i * m + j;   // 我用行存储
                    b = ti * m + tj;
                     if ( ! Bound(ti,tj))
                         continue ;
                     if (k == 0 )   // 向左
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' C ' || map[i][j] == ' F ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' I ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' B ' || map[ti][tj] == ' D ' || map[ti][tj] == ' F ' || map[ti][tj] == ' G ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                     else       // 向上
                    {
                         if (map[i][j] == ' A ' || map[i][j] == ' B ' || map[i][j] == ' E ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' J ' || map[i][j] == ' K ' )
                        {
                             if (map[ti][tj] == ' C ' || map[ti][tj] == ' D ' || map[ti][tj] == ' E ' || map[ti][tj] == ' H ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' )
                            {
                                Merge(a,b);
                            }
                        }
                    }
                }
            }
         int  sum = 0 ;
         for (i = 0 ;i < m;i ++ )
             for (j = 0 ;j < n;j ++ )
            {
                 if (un[i * m + j] == (i * m + j))
                    sum ++ ;
            }
        printf( " %d\n " ,sum);
    }
     return   0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/11/1454349.html