bzoj 2730: [HNOI2012]矿场搭建

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #define M 508
 using namespace std;
 int T=,m,n,head[M],K,next[*M],u[*M],cnt,f[M],c[M],fa[M],dfn[M],low[M],time1,ans,bi[M],sum[M];
 long long ans1=;
 void jia(int a1,int a2)
 {
     cnt++;
     next[cnt]=head[a1];
     head[a1]=cnt;
     u[cnt]=a2;
 }
 void tarjin(int a1,int deep)
 {
     time1++;
     dfn[a1]=low[a1]=time1;
     c[a1]=-;
     int tot=;
     for(int i=head[a1];i;i=next[i])
     if(u[i]!=fa[a1])
       {
         fa[u[i]]=a1;
         if(!c[u[i]])
           {
             tarjin(u[i],deep+);
             low[a1]=min(low[a1],low[u[i]]);
             tot++;
             if(deep==&&tot>)
               f[a1]=;
             if(deep>&&low[u[i]]>=dfn[a1])
               f[a1]=;
           }
         else  if(c[u[i]]==-&&u[i]!=fa[a1])
                  low[a1]=min(low[a1],dfn[u[i]]);
       }
     c[a1]=;
 }
 void dfs(int a1)
 {
     c[a1]=;
     if(!bi[a1])
       bi[a1]=time1;
     else
       bi[a1]=-;
     for(int i=head[a1];i;i=next[i])
       if(!c[u[i]]&&!f[u[i]])
         dfs(u[i]);
     return;
 }
 int main()
 {
     for(;scanf("%d",&m),m;T++){
     ans=cnt=n=;
     ans1=;
     memset(head,,sizeof(head));
     memset(bi,,sizeof(bi));
     memset(f,,sizeof(f));
     memset(sum,,sizeof(sum));
     for(int i=;i<=m;i++)
       {
         int a1,a2;
         scanf("%d%d",&a1,&a2);
         n=max(n,a1);
         n=max(n,a2);
         jia(a1,a2);
         jia(a2,a1);
       }
     time1=;
     for(int i=;i<=n;i++)
       {
          f[i]=;
          c[i]=;
          fa[i]=;
       }
     tarjin(,);
     time1=;
     for(int i=;i<=n;i++)
       if(f[i])
         {
           memset(c,,sizeof(c));
           for(int j=head[i];j;j=next[j])
             if(!c[u[j]]&&!f[u[j]])
               {
                 time1++;
                 dfs(u[j]);
               }
         }
     for(int i=;i<=n;i++)
       if(bi[i]!=-)
         sum[bi[i]]++;
     for(int i=;i<=time1;i++)
       if(sum[i])
         {
             ans++;
             ans1*=sum[i];
         }
     printf("Case %d: ",T);
     if(ans)
       printf("%d %lld\n",ans,ans1);
     else
       printf("2 %d\n",n*(n-)/);}
     return ;
 }

先用tarjin找割点 割点条件u1是树根，且有大于1棵子树，u1不是树根，low[u[i]]>dfn[u1],把和一个割点相连的联通块建出口。