动态询问连通图任意两点间最短路，单次询问.
显然，肯定有一些巧妙地性质（不然你就发明了新的最短路算法了233）
有一点很奇怪：边数最多只比点数多 $20$ 个，那么就可以将这个图看作是一个生成树，上面连了不到 $20$ 条边.
考虑两个点之间地最短路只有两种情况：经过所有只在生成树上的点，或者经过一些连着生成树外的点.
第一个情况非常好求，随便搞一个生成树然后求个距离就行.
对于第二种情况，由于连着生成树外的边的点最多只有 $20$ 个，所以可以对这些点都跑一遍最短路，然后依次枚举即可.
每次询问的复杂度为 $O(log$ $n+20 )$
#include<bits/stdc++.h>
using namespace std;
void setIO(string s) {
string in=s+".in";
freopen(in.c_str(),"r",stdin);
}
typedef long long ll;
const int maxn=100005;
const ll inf=10000000000000;
struct Union {
int p[maxn];
void init() {
for(int i=0;i<maxn;++i) p[i]=i;
}
int find(int x) {
return p[x]==x?x:p[x]=find(p[x]);
}
int merge(int x,int y) {
int a=find(x),b=find(y);
if(a==b) return 0;
p[a]=b;
return 1;
}
}ufs;
struct Edge {
int u,v,c;
}ed[maxn];
namespace tree {
int edges;
int val[maxn<<1],hd[maxn],to[maxn<<1],nex[maxn<<1];
int dep[maxn],fa[maxn],top[maxn],siz[maxn],son[maxn];
ll dis[maxn];
void addedge(int u,int v,int c) {
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
void dfs1(int u,int ff) {
fa[u]=ff,dep[u]=dep[ff]+1,siz[u]=1;
for(int i=hd[u];i;i=nex[i]) {
int v=to[i];
if(v==ff) continue;
dis[v]=dis[u]+1ll*val[i];
dfs1(v,u), siz[u]+=siz[v];
if(siz[v]>siz[son[u]]) son[u]=v;
}
}
void dfs2(int u,int tp) {
top[u]=tp;
if(son[u]) dfs2(son[u],tp);
for(int i=hd[u];i;i=nex[i]) {
int v=to[i];
if(v==fa[u]||v==son[u]) continue;
dfs2(v,v);
}
}
int LCA(int x,int y) {
while(top[x]^top[y]) {
dep[top[x]] > dep[top[y]] ? x = fa[top[x]] : y = fa[top[y]];
}
return dep[x] < dep[y] ? x : y;
}
ll Dis(int x,int y) {
return dis[x]+dis[y]-(dis[LCA(x,y)]*2);
}
};
namespace Dijkstra {
struct Node {
ll dis; int u;
Node(ll dis=0,int u=0):dis(dis),u(u){}
bool operator<(Node b) const {
return dis>b.dis;
}
};
priority_queue<Node>Q;
int edges;
int done[maxn], hd[maxn],to[maxn<<1],nex[maxn<<1],val[maxn<<1];
ll d[maxn];
ll dis[50][maxn];
void addedge(int u,int v,int c) {
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
void dijkstra(int s,int idd) {
for(int i=0;i<maxn;++i) d[i]=inf,done[i]=0;
d[s]=0, dis[idd][s]=0, Q.push(Node(0,s));
while(!Q.empty()) {
Node e=Q.top(); Q.pop();
int u=e.u;
if(done[u]) continue;
done[u]=1, dis[idd][u]=d[u];
for(int i=hd[u];i;i=nex[i]) {
int v=to[i];
if(d[u]+val[i]<d[v]) {
d[v]=d[u]+1ll*val[i];
Q.push(Node(d[v], v));
}
}
}
}
};
int n,m;
int mk[maxn],ne[maxn];
int main() {
// setIO("input");
scanf("%d%d",&n,&m);
ufs.init();
for(int i=1;i<=m;++i) {
scanf("%d%d%d",&ed[i].u,&ed[i].v,&ed[i].c);
if(ufs.merge(ed[i].u, ed[i].v)) {
tree::addedge(ed[i].u, ed[i].v, ed[i].c);
tree::addedge(ed[i].v, ed[i].u, ed[i].c);
mk[i]=1;
}
}
int cnt=0;
tree::dfs1(1,0);
tree::dfs2(1,1);
for(int i=1;i<=m;++i) if(mk[i]==0) ne[ed[i].u]=ne[ed[i].v]=1;
for(int i=1;i<=m;++i) Dijkstra::addedge(ed[i].u, ed[i].v, ed[i].c), Dijkstra::addedge(ed[i].v,ed[i].u,ed[i].c);
for(int i=1;i<=n;++i) if(ne[i]) ++cnt, Dijkstra::dijkstra(i,cnt);
int Q;
scanf("%d",&Q);
for(int cas=1;cas<=Q;++cas) {
int u,v;
scanf("%d%d",&u,&v);
ll ans=tree::Dis(u,v);
for(int i=1;i<=cnt;++i) ans=min(ans,Dijkstra::dis[i][u]+Dijkstra::dis[i][v]);
printf("%I64d\n",ans);
}
return 0;
}