和P2685 [TJOI2012]桥可以说是一模一样，判断$u,v$是否在最短路径上可以利用之前预处理的id[]详细看代码。

#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
using ll=long long;
using pii=pair<int,int>;
using pli=pair<long long,int>;
constexpr int N=200010,M=400010;
constexpr ll INF=0x3f3f3f3f3f3f3f3f;
int h[N],e[M],ne[M],w[M],idx;
void add(int a,int b,int c){e[idx]=b,ne[idx]=h[a],w[idx]=c,h[a]=idx++;}
int n,m;
bool st[N];
ll d1[N],dn[N];
int path[N],cnt,id[N];
bool in[M];
int L[N],R[N];
ll ans[N];
map<pii,int> mp;
int S,T;void dijkstra(int S,ll d[])
{memset(d,0x3f,(n+1)*sizeof(ll));memset(st,0,sizeof st);d[S]=0;priority_queue<pli,vector<pli>,greater<pli> >q;q.push({0,S});while(q.size()){int u=q.top().second;q.pop();if(st[u]) continue;st[u]=1;for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(d[v]<=d[u]+w[i]) continue;d[v]=d[u]+w[i];q.push({d[v],v});}}
}
void bfs(int k,ll d[],int f[])
{queue<int> q;q.push(path[k]);f[path[k]]=k;while(q.size()){int u=q.front();q.pop();for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(!id[v]&&!f[v]&&d[u]+w[i]==d[v]) f[v]=k,q.push(v);}}
}
struct node
{int l,r;ll v;
}tree[N<<2];
void build(int u,int l,int r)
{tree[u]={l,r,INF};if(l==r) return;int mid=l+r>>1;build(u<<1,l,mid),build(u<<1|1,mid+1,r);
}
void modify(int u,int l,int r,ll x)
{if(tree[u].l>=l&&tree[u].r<=r) return tree[u].v=min(tree[u].v,x),void();int mid=tree[u].l+tree[u].r>>1;if(l<=mid) modify(u<<1,l,r,x);if(r>mid) modify(u<<1|1,l,r,x);
}
void pushdown(int u)
{if(tree[u].l==tree[u].r){ans[tree[u].l]=tree[u].v;return;}tree[u<<1].v=min(tree[u<<1].v,tree[u].v);tree[u<<1|1].v=min(tree[u<<1|1].v,tree[u].v);int mid=tree[u].l+tree[u].r>>1;pushdown(u<<1),pushdown(u<<1|1);}
int onpath(int u,int v)
{if(!id[u]||!id[v]) return 0;if(id[u]+1==id[v]) return id[u];if(id[v]+1==id[u]) return id[v];return 0;
}
int main()
{cin>>n>>m;memset(h,-1,sizeof h);for(int i=1;i<=m;i++){int a,b,c;cin>>a>>b>>c;add(a,b,c),add(b,a,c);}cin>>S>>T;dijkstra(S,d1),dijkstra(T,dn);// 预处理1->n路径上的点int u=S;while(u!=T){path[++cnt]=u;id[u]=cnt;for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(dn[v]+w[i]==dn[u]) {in[i]=1;u=v;break;}}}path[++cnt]=T;id[T]=cnt;for(int i=1;i<=cnt;i++) bfs(i,d1,L),bfs(i,dn,R);--cnt;build(1,1,cnt);for(int u=1;u<=n;u++)for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(!in[i]&&L[u]<R[v])modify(1,L[u],R[v]-1,d1[u]+w[i]+dn[v]);}pushdown(1);int q;cin>>q;while(q--){int u,v;cin>>u>>v;ll res=onpath(u,v)?ans[onpath(u,v)]:d1[T];if(res==INF) cout<<"Infinity\n";else cout<<res<<'\n';}return 0;
}