cf1051F. The Shortest Statement（最短路/dfs树）

You are given a weighed undirected connected graph, consisting of nn vertices and mm edges.

You should answer qq queries, the ii-th query is to find the shortest distance between vertices uiui and vivi.

Input

The first line contains two integers n and m (1≤n,m≤105,m−n≤20) — the number of vertices and edges in the graph.

Next m lines contain the edges: the i-th edge is a triple of integers vi,ui,di (1≤ui,vi≤n,1≤di≤109,ui≠vi). This triple means that there is an edge between vertices ui and vi of weight di. It is guaranteed that graph contains no self-loops and multiple edges.

The next line contains a single integer q (1≤q≤105)— the number of queries.

Each of the next qq lines contains two integers uiui and vi (1≤ui,vi≤n)vi (1≤ui,vi≤n) — descriptions of the queries.

Pay attention to the restriction m−n ≤ 20

Output

Print q lines.

The i-th line should contain the answer to the i-th query — the shortest distance between vertices ui and vi.

解题思路：

题目非常良心地强调了m-n<=20

建一颗dfs树，对于未加入树的边两端跑Dij

询问只要枚举未入树边更新即可。

代码：

 #include<queue>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 typedef long long lnt;
 struct pnt{
     int hd;
     int dp;
     int no;
     int fa[];
     lnt dis;
     lnt tds;
     bool vis;
     bool cop;
     bool friend operator < (pnt x,pnt y)
     {
         return x.dis>y.dis;
     }
 }p[];
 struct ent{
     int twd;
     int lst;
     lnt vls;
 }e[];
 std::priority_queue<pnt>Q;
 int n,m;
 int q;
 int cnt;
 int sct;
 lnt dist[][];
 int sta[];
 void ade(int f,int t,lnt v)
 {
     cnt++;
     e[cnt].twd=t;
     e[cnt].vls=v;
     e[cnt].lst=p[f].hd;
     p[f].hd=cnt;
     return ;
 }
 void Dij(int x)
 {
     if(p[x].cop)
         return ;
     p[x].cop=true;
     sta[++sct]=x;
     for(int i=;i<=n;i++)
     {
         p[i].dis=0x3f3f3f3f3f3f3f3fll;
         p[i].no=i;
         p[i].vis=false;
     }
     p[x].dis=;
     while(!Q.empty())
         Q.pop();
     Q.push(p[x]);
     while(!Q.empty())
     {
         x=Q.top().no;
         Q.pop();
         if(p[x].vis)
             continue;
         p[x].vis=true;
         for(int i=p[x].hd;i;i=e[i].lst)
         {
             int to=e[i].twd;
             if(p[to].dis>p[x].dis+e[i].vls)
             {
                 p[to].dis=p[x].dis+e[i].vls;
                 Q.push(p[to]);
             }
         }
     }
     for(int i=;i<=n;i++)
         dist[sct][i]=p[i].dis;
     return ;
 }
 void dfs(int x,int f)
 {
     p[x].fa[]=f;
     p[x].dp=p[f].dp+;
     for(int i=;i<=;i++)
         p[x].fa[i]=p[p[x].fa[i-]].fa[i-];
     for(int i=p[x].hd;i;i=e[i].lst)
     {
         int to=e[i].twd;
         if(to==f)
             continue;
         if(p[to].tds)
         {
             Dij(x);
             Dij(to);
         }else{
             p[to].tds=p[x].tds+e[i].vls;
             dfs(to,x);
         }
     }
 }
 int Lca(int x,int y)
 {
     if(p[x].dp<p[y].dp)
         std::swap(x,y);
     for(int i=;i>=;i--)
         if(p[p[x].fa[i]].dp>=p[y].dp)
             x=p[x].fa[i];
     if(x==y)
         return x;
     for(int i=;i>=;i--)
         if(p[x].fa[i]!=p[y].fa[i])
         {
             x=p[x].fa[i];
             y=p[y].fa[i];
         }
     return p[x].fa[];
 }
 int main()
 {
     scanf("%d%d",&n,&m);
     for(int i=;i<=m;i++)
     {
         int a,b,c;
         scanf("%d%d%d",&a,&b,&c);
         ade(a,b,c);
         ade(b,a,c);
     }
     p[].tds=;
     dfs(,);
     scanf("%d",&q);
     while(q--)
     {
         int x,y;
         scanf("%d%d",&x,&y);
         int f=Lca(x,y);
         lnt ans=p[x].tds+p[y].tds-*p[f].tds;
         for(int i=;i<=sct;i++)
             ans=std::min(ans,dist[i][x]+dist[i][y]);
         printf("%I64d\n",ans);
     }
     return ;
 }