次小生成树模板

算法思路：求最小生成树，用maxx[i][j]来表示最小生成树中i到j的最大边权，求完后，直接枚举所有不在最小生成树中的边，替换最大边权的边，更新答案

bool vis[maxn],used[maxn][maxn];
int n,dis[maxn],pre[maxn],maxx[maxn][maxn],cost[maxn][maxn];
int Prim()
{
    int ans=inf,tot=0;
    memset ( vis , false , sizeof(vis) );
    memset ( maxx , 0 , sizeof(maxx) );
    memset ( used , false , sizeof(used) );
    vis[0] = true;
    pre[0] = -1;
    for ( int i=1 ; i<n ; i++ )
        dis[i] = cost[0][i],pre[i] = 0;
    dis[0] = 0;
    while ( true )
    {
        int minc = inf,p = -1;
        for ( int i=0 ; i<n ; i++ )
            if ( !vis[i]&&dis[i]<minc )
                minc = dis[i],p = i;
        if ( p==-1 ) break;
        tot += minc;
        vis[p] = true;
        used[p][pre[p]] = used[pre[p]][p] = true;
        for ( int i=0 ; i<n ; i++ )
        {
            if ( vis[i]&&i!=p )
                maxx[i][p] = maxx[p][i] = Max( maxx[i][pre[p]] , dis[p] );
            if ( !vis[i]&&dis[i]>cost[p][i] )
                dis[i] = cost[p][i],pre[i] = p;
        }
    }
    for ( int i=0 ; i<n ; i++ )
        for ( int j=i+1 ; j<n ; j++ )
            if ( !used[i][j] ) ans = Min ( ans , tot-maxx[i][j]+cost[i][j] );
    return ans;
}