HDUOJ--------（1198）Farm Irrigation

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4491    Accepted Submission(s): 1947

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 1Benny 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 likeFigure 2Several 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

 

Source

Zhejiang University Local Contest 2005

 

Recommend

Ignatius.L

 

代码：

/*龚细军 宽度优先搜索bfs*/
#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
typedef struct G
{
 /*true 代表此处有piper*/
 bool up,down,left,right;
}gong; 
struct point
{
    int x;
    int y;
}start;
 //依次代表A~K的管道特性;
 gong go[]={1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,1,0,0,0,0,1,1,1,0,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1};
 
 /*前行的步奏*/
 char map[51][51];
 int n,ans,m;
void input()
{
  memset(map,'',sizeof(map));
  for(int i=0;i<m;i++)
      scanf("%s",map[i]);
}
void dfs()
{
    /*判断是否联通，这样便于构造重位*/
    bool isrr,isll,isuu,isdd;
    queue<point> q;
    point p1,p2;
    q.push(start);
    bool isbian=false;
    while(!q.empty())
    {
        isrr=isll=isuu=isdd=false;
        p1=q.front();
        q.pop();
        /*向下搜索*/
        if(p1.x+1<m&&map[p1.x+1][p1.y]>='A'&&go[map[p1.x][p1.y]-'A'].down!=0&&go[map[p1.x+1][p1.y]-'A'].up!=0)
        {
            p2.x=p1.x+1;
            p2.y=p1.y;
            q.push(p2);
            isdd=true; //说明经过这里；            
        }
        /*向上搜索*/
        if(p1.x>=1&&map[p1.x-1][p1.y]>='A'&&go[map[p1.x][p1.y]-'A'].up!=0&&go[map[p1.x-1][p1.y]-'A'].down!=0)
        {
          p2.x=p1.x-1;
          p2.y=p1.y;
          q.push(p2);
          isuu=true;
        }
        /*向左搜索*/
        if(p1.y>=1&&map[p1.x][p1.y-1]>='A'&&go[map[p1.x][p1.y]-'A'].left!=0&&go[map[p1.x][p1.y-1]-'A'].right!=0)
        {
            p2.x=p1.x;
            p2.y=p1.y-1;
            q.push(p2);
            isll=true;
        }
        /*向右搜索*/
        if(p1.y+1<n&&map[p1.x][p1.y+1]>='A'&&go[map[p1.x][p1.y]-'A'].right!=0&&go[map[p1.x][p1.y+1]-'A'].left!=0)
        {
            p2.x=p1.x;
            p2.y=p1.y+1;
            q.push(p2);
            isrr=true;
        }
        if(isuu||isll||isrr||isdd)
        {
            map[p1.x][p1.y]='0';
        }
      else 
            if((p1.x>=1&&map[p1.x-1][p1.y]=='0')||(p1.x+1<m&&map[p1.x+1][p1.y]=='0')
              ||(p1.y>=1&&map[p1.x][p1.y-1]=='0')||(p1.y+1<n&&map[p1.x][p1.y+1]=='0'))
          {
              if(map[p1.x][p1.y]!='0')
              {
               isbian=true;
               map[p1.x][p1.y]='1';
              }
          }
      
        else ans++;
    }
    if(isbian) ans++;
}
int main()
{
    int i,j;
   
   /* for(i=0;i<11;i++)
    {
        printf("%d %d %d %d
",go[i].up,go[i].down,go[i].left,go[i].right);
    }*/
    
    while(scanf("%d%d",&m,&n),(m!=-1||n!=-1))
    {
      ans=0;
      input( );
     for(i=0;i<m;i++)
     {
         for(j=0;j<n;j++)
         {
             if(map[i][j]>='A')
             {
              start.x=i;
              start.y=j;
              dfs();
             }
         }
     }
     printf("%d
",ans);
    }
    return 0;  
}