判断最小生成树是否唯一

判断最小生成树是否唯一

http://202.120.80.191/problem.php?problemid=2614

#include  < iostream >
#include  < algorithm >
#include  < vector >
#include  < string >
using   namespace  std;
const   int  maxn = 110 ;
int  g[maxn][maxn],pre[maxn];
int  maxbetween[maxn][maxn];
int  x[maxn],y[maxn];
int  n;
int  ans;
void  init()
{
     for ( int  i = 0 ;i < n;i ++ )
    {
        g[i][i] = 0 ;
         for ( int  j = i + 1 ;j < n;j ++ )
            g[i][j] = g[j][i] = abs(x[i] - x[j]) + abs(y[i] - y[j]);
    }
}
int  prim( int  u)
{
     int  ans = 0 ;
     int  k,smin;
     int  dist[maxn];
     bool  visit[maxn];
    memset(maxbetween, 0 , sizeof (maxbetween));
     for ( int  i = 0 ;i < n;i ++ )
    {
        dist[i] = g[u][i];
        visit[i] = 0 ;
        pre[i] = u;
    }
    visit[u] = 1 ;
    pre[u] =- 1 ;
     for ( int  i = 1 ;i < n;i ++ )
    {
        k =- 1 ;
        smin = INT_MAX;
         for ( int  j = 0 ;j < n;j ++ )
             if ( ! visit[j]  &&  dist[j] < smin)
            {
                k = j;
                smin = dist[j];
            }
         // if(k==-1) continue;
        
         for ( int  j = 0 ;j < n;j ++ )
             if (visit[j])
                maxbetween[j][k] = maxbetween[k][j] = max(maxbetween[j][pre[k]],smin);
        
        ans += smin;
        visit[k] = 1 ;

         for ( int  j = 0 ;j < n;j ++ )
             if ( ! visit[j]  &&  g[k][j] < dist[j])
            {
                dist[j] = g[k][j];
                pre[j] = k;
            }
    }
     return  ans;
}
int  unique()
{
     int  res = INT_MAX;
     for ( int  i = 0 ;i < n;i ++ )
         for ( int  j = 0 ;j < n;j ++ )
             if (i != j  &&  i != pre[j]  &&  j != pre[i] )
                res = min(res,ans - maxbetween[i][j] + g[i][j]);
     return  res;
}
int  main()
{
     while (scanf( " %d " , & n) != EOF)
    {
         for ( int  i = 0 ;i < n;i ++ )
            scanf( " %d%d " , & x[i], & y[i]);
        init();
        ans = prim( 0 );
        printf( " %d\n%s\n " ,ans,unique() == ans ? " Yes " : " No " );
    }
}