luogu P3576 [POI2014]MRO-Ant colony

Posted 2021-01-07 smyjr

传送门

一群蚂蚁能被吃,也就是走到指定边的两端点之一要走到另一端点时有(k)只,我们可以从这两端点逆推,记两个值为走到某个点时最后会被吃掉(k)只蚂蚁的蚂蚁数量范围,式子下面有,很好理解(雾).最后在每个叶子节点二分查找有多少个数在区间内即可

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define LL long long
#define il inline
#define re register
#define inf 2099999999
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define db double
#define eps (1e-5)

using namespace std;
const int N=1000000+10;
il LL rd()
{
    re LL x=0,w=1;re char ch=0;
    while(ch<'0'||ch>'9') {if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    return x*w;
}
LL d[N];
int to[N<<1],nt[N<<1],hd[N],tot=1;
il void add(int x,int y)
{
  ++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot,++d[x];
  ++tot,to[tot]=x,nt[tot]=hd[y],hd[y]=tot,++d[y];
}
int n,g,k,ma,xx,yy;
LL ll[N],rr[N],ans,a[N];
void dfs(int x,int ffa)
{
  for(int i=hd[x];i;i=nt[i])
    {
      int y=to[i];
      if(y==ffa) continue;
      ll[y]=min(ll[x]*max(d[x]-1,1ll),a[g]+10),rr[y]=min((rr[x]+1)*max(d[x]-1,1ll)-1,a[g]+10);
      dfs(y,x);
    }
  if(d[x]==1)
    {
      int l=0,r=g,z=g,zz=0;
      while(l<=r)
        {
          int mid=(l+r)>>1;
          if(a[mid]<ll[x]) z=mid,l=mid+1;
          else r=mid-1;
        }
      l=0,r=g;
      while(l<=r)
        {
          int mid=(l+r)>>1;
          if(a[mid]<=rr[x]) zz=mid,l=mid+1;
          else r=mid-1;
        }
      ans+=max(zz-z,0);
    }
}

int main()
{
  n=rd(),g=rd(),k=rd();
  for(int i=1;i<=g;i++) a[i]=rd();
  sort(a+1,a+g+1);
  a[0]=-1;
  add(xx=rd(),yy=rd());
  for(int i=2;i<n;i++) add(rd(),rd());
  ll[xx]=rr[xx]=k,dfs(xx,yy);
  ll[yy]=rr[yy]=k,dfs(yy,xx);
  printf("%lld
",ans*k);
  return 0;
}