poj 3026 Borg Maze

道题比较坑，就是输入N,M之后可能有多个空格这里要注意；

方法：1：建图，把A进行编号在存储到图中；

2：利用BFS找出任意两点的最短距离；

3：利用kruscal生成最小生成树；

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<vector>
using namespace std;
int d[4][2] = { 0,1,1,0,0,-1,-1,0 };
class Node
{
public:
      int x,y,step;
}q[100024];
class node
{
public:
       int x,y;    
}e[124];
Node kru[20024];
int set[124];
bool cmp( Node a ,Node b  )
{
    return a.step < b.step;    
}
int find( int x )
{
   return set[x] == x ? x : set[x] = find( set[x] );     
}
int Kruscal( int N )
{
  int ans=0,X,Y;
  for( int i = 0 ; i < N ; i++ )
  {
     if( ( X = find( kru[i].x ) )!=( Y = find(kru[i].y) ) )
     {
         ans += kru[i].step;
         set[Y] =X;        
     }
  }    
  return ans;
}
int BFS( int x,int y , int end ,int map[][54])
{
    int S=0,E=0;
    bool hash[54][54] = {0};
    hash[x][y] = true;;
    q[S].x = x ;q[S].y = y; q[S++].step = 0;
    while( E < S )
    {
        Node p = q[E];
        for( int i = 0 ; i < 4 ; i ++ )    
        {
            int dx = d[i][0] + p.x ,dy = d[i][1] + p.y;
            if( !hash[dx][dy] && map[dx][dy] != -1 )
            {
                if( map[dx][dy]==end ) return p.step + 1;
                q[S].x = dx , q[S].y = dy;
                q[S++].step = p.step + 1; 
                hash[dx][dy] = true; 
            }    
        }
        E++;
    }
    return 0x7fffffff;
}
int main(  )
{
    int map[54][54],Case,N,M;
    char str[54],c;
    while( scanf( "%d",&Case )==1 )
    {
       while( Case-- )
       {
          scanf( "%d %d",&N,&M );
          int count = 2,cnt=0;
          for( int i = 0 ; i <54 ; i++ )
          {
              for( int j = 0 ; j < 54 ; j ++ )
                 map[i][j]  = -1;        
          }
          while( c = getchar(  ),c!='\n' );
          for( int i = 1 ; i <= M ; i ++ )//建图 
          {              
               gets(str);
               for( int j = 0 ; j < N ; j ++ )
               {
                  if( str[j] =='#' )
                      map[i][j+1] = -1;
                  else 
                  if( str[j] ==' ' ) 
                           map[i][j+1] = 0;
                 else  
                   if( str[j] =='S' ) 
                   {
                     map[i][j+1] = 1;e[1].x = i ;e[1].y = j+1;
                   } 
                 else 
                 {
                    e[count].x = i ;e[count].y = j+1;
                    map[i][j+1] = count++;
                 }        
               }        
          }
          for( int i = 1 ; i < count ; i ++ )//搜索出i点到任意点的最短距离 
          {
             set[i] = i;
             for( int j = i + 1 ; j < count ; j ++ )
             {
               kru[cnt].x = i;kru[cnt].y = j;
               kru[cnt++].step = BFS( e[i].x ,e[i].y , j  ,map);
             }            
          }
          sort( kru ,kru + cnt , cmp );    
          printf( "%d\n",Kruscal( cnt ) );    
       }    
    }
    //system( "pause" );
    return 0;
}