#pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include<stdio.h> #include<string.h> #include<vector> #include<algorithm> #include<cmath> #include<map> #include<queue> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1 | 1 #define lowbit(x) x&(-x) template <class T> inline bool rd(T &ret){ char c;int sgn; if(c=getchar(),c==EOF) return 0; while(c!='-' && (c<'0' || c>'9')) c=getchar(); sgn=(c=='-')?-1:1; ret=(c=='-')?0:(c-'0'); while(c=getchar(),c>='0' && c<='9') ret=ret*10+(c-'0'); ret*=sgn; return 1; } template <class T> inline void pt(T x){ if(x<0){ putchar('-'); x=-x; } if(x>9) pt(x/10); putchar(x%10+'0'); } using namespace std; typedef long long ll; const int N=3005; long long inf=1e15; const int MOD=1e9+7; int T,n,m,k,x,y,z,l,tot,cnt; ll g[N][N],d[N],mst,cost[N][N]; bool vis[N],choose[N][N]; vector<int> G[N]; int pre[N]; ll dfs(int u,int fa,int src){/*返回src为根的树中,src到u子树的非树边的最短距离*/ ll siz=inf; for(int i=0;i<G[u].size();i++){ int v=G[u][i]; if(v==fa) continue; ll tmp=dfs(v,u,src); siz=min(siz,tmp); cost[u][v]=min(cost[u][v],tmp); } if(fa!=src)/*如果不是树边*/ siz=min(siz,g[u][src]); return siz; } void MST(){ for(int i=0;i<n;i++) for(int j=0;j<n;j++) cost[i][j]=g[i][j]=inf,choose[i][j]=0; while(m--){ int u,v;ll dis;rd(u);rd(v);rd(dis); g[u][v]=g[v][u]=min(g[u][v],dis); } for(int i=0;i<n;i++){ d[i]=inf; G[i].clear(); vis[i]=0; pre[i]=-1; } d[0]=0; mst=0; for(int i=0;i<n;i++){ int pos=-1; for(int j=0;j<n;j++) if(!vis[j] && (pos==-1 || d[pos]>d[j])) pos=j; if(pre[pos]!=-1){ G[pos].push_back(pre[pos]); G[pre[pos]].push_back(pos); choose[pos][pre[pos]]=choose[pre[pos]][pos]=1; } for(int j=0;j<n;j++) if(d[j]>g[j][pos]){ d[j]=g[j][pos]; pre[j]=pos; } vis[pos]=1; mst+=d[pos]; } } int main() { #ifndef ONLINE_JUDGE freopen("aaa","r",stdin); #endif int q,u,v;ll dis; while(~scanf("%d%d",&n,&m),m+n){ MST(); for(int i=0;i<n;i++) dfs(i,-1,i); rd(q); ll ans=0; for(int i=1;i<=q;i++){ rd(u);rd(v);rd(dis); if(choose[u][v]==false) ans+=mst;/*修改边不在最小生成树上*/ else ans+=mst-g[u][v]+min(cost[u][v],dis);/*修改边在最小生成树上*/ } printf("%.4f\n",(double)ans/(double)q); } return 0; }