USACO 2.4 Bessie Come Home

USACO 2.4 Bessie Come Home

最短路径问题，数据量很小(最多52个结点)，所以用floyd算法就可以了。

#include  < iostream >
#include  < fstream >

using   namespace  std;

ifstream fin( " comehome.in " );
ofstream fout( " comehome.out " );

#ifdef _DEBUG
#define  out cout
#define  in cin
#else
#define  out fout
#define  in fin
#endif

int  graph[ 52 ][ 52 ];

int  char2int( char  c)
{
     if (c >= ' a ' && c <= ' z ' )
         return  c - ' a ' ;
     else   if (c >= ' A ' && c <= ' Z ' )
         return  c - ' A ' + 26 ;
}

void  floyd()
{
     for ( int  k = 0 ;k < 52 ; ++ k)
         for ( int  i = 0 ;i < 52 ; ++ i)
             for ( int  j = 0 ;j < 52 ; ++ j){
                graph[i][j]  =  min(graph[i][j],graph[i][k] + graph[k][j]);
            }
}

void  solve()
{
     int  p;
     char  a,b;
     int  len;

     for ( int  i = 0 ;i < 52 ; ++ i)
         for ( int  j = 0 ;j < 52 ; ++ j)
            graph[i][j]  =  INT_MAX / 2 ;

     in >> p;

     while (p -- ){
        in >> a >> b >> len;
        int  i  =  char2int(a);
        int  j  =  char2int(b);
       graph[i][j]  =  graph[j][i]  =  min(graph[i][j],len);
    }

    floyd();

     int  res  =  INT_MAX;
     int  z  =  char2int( ' Z ' );
     char  resi;

     for ( int  i = 26 ;i < 51 ; ++ i){
         if (graph[z][i] < res){
            resi  =   ' A ' + i - 26 ;
            res  =  graph[z][i];
        }
    }

     out << resi << "   " << res << endl;
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}