HDU1162最小生成树MST

HDU1162最小生成树MST

求MST两种算法：Prim和Kruskal。Prim以点作为主体，Kruskal以边作为主体。

#include  < iostream >
#include  < cmath >

using   namespace  std;


const   int  M = 101 ;
const   double  MAX = 10000000 ;

struct  S
{
     double  x,y;
}data[M];

double  map[M][M];
double  dis[M];  // 存储i节点到当前节点的最短距离
int  n;
double  res;  // 结果

double  d(S a,S b)
{
     return  sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

void  prim( int  v)
{
     int  i,j,t;
     double  min;
     for (i = 1 ;i <= n;i ++ )
        dis[i] = map[v][i];
    dis[v] = 0 ;         // 标记已访问
    res = 0 ;
     for (t = 1 ;t < n;t ++ )  // 找点
    {
        min = MAX;
         for (i = 1 ;i <= n;i ++ )
             if (dis[i] < min  &&  dis[i] != 0 )  // 找到未访问过的到当前节点最短的点
            {
                min = dis[i];
                j = i;
            }
        res += min;
        dis[j] = 0 ;
         for (i = 1 ;i <= n;i ++ )             // 更新dis数组
             if (map[j][i] < dis[i])
                dis[i] = map[j][i];
    }
}

int  main()
{
     int  i,j;
     while (scanf( " %d " , & n) != EOF)
    {
         for (i = 1 ;i <= n;i ++ )
            scanf( " %lf %lf " , & data[i].x, & data[i].y);
         for (i = 1 ;i <= n;i ++ )
             for (j = 1 ;j <= n;j ++ )
                map[i][j] = d(data[i],data[j]);
        prim( 1 );
        printf( " %.2lf\n " ,res);
    }
     return   0 ;
}