解题：POI 2014 Ant colony

题面

既然我们只知道最后数量为$k$的蚂蚁会在特殊边上被吃掉，不妨逆着推回去，然后到达每个叶节点的时候就会有一个被吃掉的蚂蚁的区间，然后二分一下就好啦

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,maxx=1e9;
 int ant[N],deg[N],leaf[N];
 int p[N],noww[*N],goal[*N];
 int n,g,k,t,root,t1,t2,n1,n2,cnt;
 long long ans;
 void link(int f,int t)
 {
     noww[++cnt]=p[f];
     goal[cnt]=t,p[f]=cnt;
 }
 void DFS(int nde,int fth,long long l,long long r)
 {
     if(l>maxx) return ; if(r>maxx) r=maxx;
     for(int i=p[nde];i;i=noww[i])
         if(goal[i]!=fth)
             t=deg[goal[i]]-,DFS(goal[i],nde,l*t,r*t+t-);
     if(leaf[nde])
         ans+=(upper_bound(ant+,ant++g,r)-lower_bound(ant+,ant++g,l))*k;
 }
 int main ()
 {
     scanf("%d%d%d",&n,&g,&k);
     for(int i=;i<=g;i++)
         scanf("%d",&ant[i]);
     sort(ant+,ant++g);
     scanf("%d%d",&n1,&n2);
     link(n1,n2),link(n2,n1);
     deg[n1]++,deg[n2]++;
     for(int i=;i<n;i++)
     {
         scanf("%d%d",&t1,&t2);
         link(t1,t2),link(t2,t1);
         deg[t1]++,deg[t2]++;
     }
     for(int i=;i<=n;i++) if(deg[i]==) deg[i]=,leaf[i]=true;
     root=n1,DFS(n1,n2,k*(deg[n1]-),k*(deg[n1]-)+deg[n1]-);
     root=n2,DFS(n2,n1,k*(deg[n2]-),k*(deg[n2]-)+deg[n2]-);
     printf("%lld",ans);
     return ;
 }