ZOJ 1203 Swordfish

ZOJ 1203 Swordfish

题目大意：给定平面上N个城市的位置，计算连接这N个城市所需线路长度总和的最小值。
MST问题。
以下是我的代码，采用Prim算法求解：

#include < cstdio >
#include < cmath >
using   namespace  std;
const   int  kMaxn( 107 );
const   int  kInf( 0x7f7f7f7f );

int  n;
double  x[kMaxn],y[kMaxn],g[kMaxn][kMaxn];
double  mst,lowcost[kMaxn];

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     int  T( 0 );
     while (scanf( " %d " , & n) == 1   &&  n)
    {
         for ( int  i = 1 ;i <= n;i ++ )
            scanf( " %lf%lf " , & x[i], & y[i]);
         for ( int  i = 1 ;i <= n;i ++ )
             for ( int  j = 1 ;j <= n;j ++ )
                g[i][j] = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));

        mst = . 0 ;
         for ( int  i = 1 ;i <= n;i ++ )
            lowcost[i] = g[ 1 ][i];
        lowcost[ 1 ] =- 1 ;
         for ( int  i = 1 ;i < n;i ++ )
        {
             int  pos( - 1 );
             double  minvalue(kInf + . 0 );
             for ( int  j = 1 ;j <= n;j ++ )
                 if (lowcost[j] !=- 1   &&  minvalue > lowcost[j])
                {
                    pos = j;
                    minvalue = lowcost[j];
                }
             if (pos !=- 1 )
            {
                mst += minvalue;
                lowcost[pos] =- 1 ;
                 for ( int  j = 1 ;j <= n;j ++ )
                     if (lowcost[j] !=- 1   &&  lowcost[j] > g[pos][j])
                        lowcost[j] = g[pos][j];
            }
        }

        T ++ ;
         if (T != 1 )
            printf( " \n " );
        printf( " Case #%d:\n " ,T);
        printf( " The minimal distance is: %.2f\n " ,mst);
    }

     return   0 ;
}