codeforces 1051 F

多源点最短路。

但是有限制，m - n <= 20，边数 - 点数 <= 20， 所以这个图非常的稀疏。

任意提取出一个生成树出来，LCA处理任意两点最短路。

然后再去遍历那些多余出来的点（非树边的两个端点），看他们能不能更新答案（相当于松弛）。这里跑40个最短路预处理出来（最多40个点，但是必须在原图上跑才行）。

 #include <bits/stdc++.h>

 using namespace std;

 #define time _____time
 const long long inf = 1E18;
 struct edge {
     long long u,v,val,next;
 };

 edge Edge[];
 edge xdge[];

 int e = ;
 int head[];

 void addedge(int u,int v,int val) {
     Edge[e].u = u;
     Edge[e].v = v;
     Edge[e].val = val;
     Edge[e].next = head[u];
     head[u] = e++;

     Edge[e].u = v;
     Edge[e].v = u;
     Edge[e].val = val;
     Edge[e].next = head[v];
     head[v] = e++;

 }

 int ec = ;
 int head2[];

 void ADD(int u,int v,int val) {
     xdge[ec].u = u;
     xdge[ec].v = v;
     xdge[ec].val = val;
     xdge[ec].next = head2[u];
     head2[u] = ec++;

     xdge[ec].u = v;
     xdge[ec].v = u;
     xdge[ec].val = val;
     xdge[ec].next = head2[v];
     head2[v] = ec++;
 }

 int in[];
 int out[];
 int time = ;
 long long depth[];
 long long ST[*][];
 void dfs(int x,int pre)
 {
     in[x]=++time;
     ST[time][]=x;
     for (int i = head2[x];i != -;i = xdge[i].next)
         if (xdge[i].v != pre){
             depth[xdge[i].v] = depth[x] + xdge[i].val;
             dfs(xdge[i].v,x);
             ST[++time][]=x;
         }
     out[x]=time;
 }

 int n,m;

 void Get_ST(int n){
     for (int i=;i<=n;i++)
         for (int j=;j<;j++){
             ST[i][j]=ST[i][j-];
             int v=i-(<<(j-));
             if (v>&&depth[ST[v][j-]]<depth[ST[i][j]])
                 ST[i][j]=ST[v][j-];
         }
 }

 int RMQ(int L,int R){
     if(L > R) {
         swap(L,R);
     }
     int val=floor(log(R-L+)/log());
     int x=ST[L+(<<val)-][val],y=ST[R][val];
     if (depth[x]<depth[y])
         return x;
     else
         return y;
 }

 int fa[];
 int ffind(int x) {
     if(fa[x] == x) return x;
     else return fa[x] = ffind(fa[x]);
 }
 void unit(int x,int y) {
     int fx,fy;
     fx = ffind(x);
     fy = ffind(y);
     if(fx != fy) {
         fa[fx] = fy;
     }
 }

 struct bian {
     int l,r,val;
 };

 vector<bian> duo;

 long long mmap[][];
 map<int,int> mp;
 vector<int> dian;

 long long dis[][];
 int vis[];

 typedef pair<long long,int> pli;

 void dij(int id,int s)
 {
     for(int i=;i<=n+;i++){
         vis[i]=;
         dis[id][i]=inf;
     }
     dis[id][s]=;
     priority_queue<pli,vector<pli>,greater<pli> >q;
     q.push(pli(,s));

     while(!q.empty())
     {
         pli tmp=q.top();
         q.pop();
         int u=tmp.second;
         if(tmp.first>dis[id][u]) continue;
         vis[u]=;

         for(int i=head[u];i!=-;i = Edge[i].next){
             int v=Edge[i].v;
             long long w=Edge[i].val;
             if(vis[v]) continue;
             if(dis[id][v]>dis[id][u]+w)
             {
                 dis[id][v]=dis[id][u]+w;
                 q.push(pli(dis[id][v],v));
             }
         }
     }

 }

 int main() {

     scanf("%d %d",&n,&m);
     int u,v;
     for(int i = ; i <= n; i++) {
         fa[i] = i;
         head[i] = -;
         head2[i] = -;
     }
     bian xx;
     int val;
     for(int i = ; i <= m; i++) {
         scanf("%d %d %d",&u,&v,&val);
         addedge(u,v,val);
         if(ffind(u) != ffind(v)) {
             unit(u,v);
             ADD(u,v,val);
         }
         else {
             dian.push_back(u);
             dian.push_back(v);
         }
     }

     sort(dian.begin(),dian.end());
     dian.erase(unique(dian.begin(),dian.end()),dian.end());

     for(int i=;i<dian.size();i++){
         //cout<<dian[i]<<endl;
         dij(i + ,dian[i]);
     }

     dfs(,);
     depth[]= inf;

     Get_ST(time);

     int q;
     scanf("%d",&q);
     while (q--){

         int x,y;
         scanf("%d %d",&x,&y);
         int LCA=RMQ(in[x],in[y]);
         long long ans = depth[x]+depth[y]-depth[LCA]*;
         for(int i = ; i < dian.size(); i++) {
             ans = min(ans,dis[i + ][x] + dis[i + ][y]);
         }

         printf("%lld\n",ans);

     }

 }