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；

View Code

#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;

}