解题：APIO 2018 铁人两项

题面

建立圆方树，考虑所有路径，发现路径上原来的点双(现在的方点)里的点都可以做中间点。但是路径上被方点夹着的圆点被计重了，要扣掉；枚举的两个端点也被算进去了，要扣掉。所以直接将方点权值设为点双大小，圆点权值设为-1，所有圆点间路径权值和的和即为答案

注意图可能不连通，草(汉语)

 #include<queue>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,M=;
 int n,m,r,c,t1,t2,cnt,Cnt,tot,top; long long sum,ans;
 int dfn[N],low[N],col[N],isc[N],stk[N],siz[N],dis[N];
 int p[N],noww[M],goal[M],P[N],Noww[M],Goal[M],val[N];
 vector<int> ve[N];
 void Link(int f,int t)
 {
     noww[++cnt]=p[f];
     goal[cnt]=t,p[f]=cnt;
     noww[++cnt]=p[t];
     goal[cnt]=f,p[t]=cnt;
 }
 void Linka(int f,int t)
 {
     Noww[++Cnt]=P[f];
     Goal[Cnt]=t,P[f]=Cnt;
     Noww[++Cnt]=P[t];
     Goal[Cnt]=f,P[t]=Cnt;
 }
 void Tarjan_PBC(int nde)
 {
     int tep=; stk[++top]=nde;
     dfn[nde]=low[nde]=++tot;
     for(int i=p[nde];i;i=noww[i])
         if(!dfn[goal[i]])
         {
             Tarjan_PBC(goal[i]);
             low[nde]=min(low[nde],low[goal[i]]);
             if(dfn[nde]<=low[goal[i]])
             {
                 if(nde!=r||++tep>) isc[nde]=true;
                 int tmp; c++;
                 do
                 {
                     tmp=stk[top--],col[tmp]=c;
                     ve[c].push_back(tmp);
                 }while(tmp!=goal[i]);
                 ve[c].push_back(nde);
             }
         }
         else low[nde]=min(low[nde],dfn[goal[i]]);
 }
 void DFS(int nde,int fth)
 {
     if(nde<=n) siz[nde]=;
     for(int i=P[nde];i;i=Noww[i])
         if(Goal[i]!=fth) DFS(Goal[i],nde),siz[nde]+=siz[Goal[i]];
 }
 void Getans(int nde,int fth)
 {
     long long tmp=;
     int sizz=tot-siz[nde];
     for(int i=P[nde];i;i=Noww[i])
         if(Goal[i]!=fth)
         {
             int G=Goal[i],S=siz[G];
             tmp+=1ll*sizz*S,sizz+=S;
             Getans(Goal[i],nde);
         }
     if(nde<=n) ans-=(tmp+tot-)*;
     else ans+=*tmp*val[nde];
 }
 int main()
 {
     scanf("%d%d",&n,&m);
     for(int i=;i<=m;i++)
         scanf("%d%d",&t1,&t2),Link(t1,t2);
     for(int i=;i<=n;val[i]=-,i++)
         if(!dfn[i]) r=i,Tarjan_PBC(i);
     for(int i=;i<=c;i++)
     {
         val[n+i]=ve[i].size();
         for(int j=;j<val[n+i];j++)
             Linka(n+i,ve[i][j]);
     }
     for(int i=;i<=n;i++)
         if(!siz[i]) DFS(i,),tot=siz[i],Getans(i,);
     printf("%lld",ans);
     return ;
 }