POJ 1751 Highways

最小生成树的水题，直接的最小生成树算法就可以了。题意给定n个城市，只不过是又给定了已经连接起来的道路，由于输出的是城镇的编号，所以不用计算两城镇之间的真正距离（既不用算sqrt）,用的是kruskal，比较费时，估计用prim可以节省不少时间.

 

code

#include  < iostream >
#include  < cstdio >
#include  < cmath >
#include  < algorithm >
#include  < memory.h >
using   namespace  std;
const   int  MAXN  =   800 ;
const   int  MAXE  =   500000 ;
const   int  INF  =   0x7fffffff ;
int  nv,ne,index, set [MAXN],pos[MAXE];
struct  Edge{
     int  u,v;
     int  val;
}edge[MAXE];
struct  Node{
     int  x,y;
}node[MAXN];
int  cmp( const  Edge &  a, const  Edge &  b)
{
     return  a.val  <  b.val;
}
int  find_set( int  x)
{
     int  root  =  x,tmp;
     while ( set [root]  >=   0 )
        root  =   set [root];
     while ( x  !=  root)
    {
        tmp  =   set [x];
         set [x]  =  root;
        x  =  tmp;
    }
     return  root;
}
void  union_set( const   int &  root1, const   int &  root2)
{
     int  a  =   set [root1];
     int  b  =   set [root1];
     if ( a > b)
    {
         set [root1]  =  root2;
         set [root2]  +=  a;
    } else {
         set [root2]  =  root1;
         set [root1]  +=  b;
    }
}
int  kruskal()
{
    index  =   0 ;
     int  sum  =   0 ,root1,root2,tmp,i;
     for (i  =   0 ;i  <  ne;  ++ i)
    {
        root1  =  find_set(edge[i].u);
        root2  =  find_set(edge[i].v);
         if ( root1  !=  root2)
        {
            pos[index]  =  i;
             ++ index;
            union_set(root1,root2);
        }
    }
     return  sum;
}
int  main()
{
     int  i,j,k,a,b,tmp,val,n,m,root1,root2;
     while (scanf( " %d " , & nv)  !=  EOF)
    {
        memset( set , - 1 , sizeof ( set ));
         for (i  =   1 ;i  <=  nv;  ++ i)
            scanf( " %d%d " , & (node[i]).x, & (node[i].y));
        index  =   0 ;
         for (i  =   1 ;i  <=  nv;  ++ i)
             for (j  =  i  +   1 ;j  <=  nv;  ++ j)
            {
                edge[index].u  =  i;
                edge[index].v  =  j;
                edge[index].val  =  (node[i].x - node[j].x) * (node[i].x - node[j].x) +
                    (node[i].y  -  node[j].y) * (node[i].y - node[j].y);
                 ++ index;
            }
        ne  =  index;
        sort(edge,edge + ne,cmp);
        scanf( " %d " , & n);
         for (i  =   1 ;i  <=  n;  ++ i)
        {
            scanf( " %d%d " , & a, & b);
            root1  =  find_set(a);
            root2  =  find_set(b);
             if ( root1  !=  root2)
                union_set(root1,root2);
        }
        kruskal();
         for (i  =   0 ;i  <  index;  ++ i)
            printf( " %d %d\n " ,edge[pos[i]].u,edge[pos[i]].v);
    }
     return   0 ;
}