2016 ACM/ICPC Dalian Online-1010 Weak Pair
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题意:给定一个有根树和k,每个节点权值为ai,求有序点对(u,v)的数量,有序点对需满足:u是v的祖先,且au*av<=k;
题解一:dfs序+主席树
求出树的dfs序,因为一棵子树的所有子节点在dfs序中是连续的,设起始序号为st,结尾序号为ed。那么对于子树的根节点i来说,只需要求[st,ed]这段区间中<=k/a[i]的个数即可,这可以用主席树实现(i点是作为祖先节点统计数量)
ps.处理时要用离散化
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>#include <set>#include <vector>using namespace std;typedef long long LL;const int N=100000+10;struct Tree{ int l,r,s;}T[5000000];int n,m,S;LL K;vector<int>V[N];int st[N],ed[N],Time;int a[N],b[N],d[N],Root[N];int cnt;void WZJ(){ for (int i=1;i<=n;i++)b[i]=a[i]; sort(b+1,b+n+1); m=unique(b+1,b+n+1)-(b+1);// for (int i=1;i<=n;i++)a[i]=lower_bound(b+1,b+m+1,a[i])-b;//}void Insert(int&p,int L,int R,int v){ cnt++;T[cnt]=T[p];p=cnt; T[p].s++; if (L==R)return ; int mid=(L+R)>>1; if (v<=mid)Insert(T[p].l,L,mid,v); else Insert(T[p].r,mid+1,R,v);}int Find(int&p,int L,int R,int l,int r){ if (l>r)return 0; if (L==l && R==r)return T[p].s; int mid=(L+R)>>1; if (r<=mid)return Find(T[p].l,L,mid,l,r);else if (mid< l)return Find(T[p].r,mid+1,R,l,r);else{ return Find(T[p].l,L,mid,l,mid)+Find(T[p].r,mid+1,R,mid+1,r); }}int GET(LL v){ if (b[1]>v)return 0; int L=1,R=m,mid,Ans; while (L<=R){ mid=(L+R)>>1; if (b[mid]<=v)Ans=mid,L=mid+1; else R=mid-1; } return Ans;}void dfs(int u){ Time++;st[u]=Time; Root[Time]=Root[Time-1]; Insert(Root[Time],1,m,a[u]); for (int i=0;i<int(V[u].size());i++){ dfs(V[u][i]); } ed[u]=Time;}void work(){ scanf("%d",&n);cin>>K; for (int i=1;i<=n;i++)scanf("%d",&a[i]),V[i].clear(); memset(d,0,sizeof d); for (int i=1;i<n;i++){ int u,v;scanf("%d%d",&u,&v); V[u].push_back(v); d[v]++; } WZJ(); for (int i=1;i<=n;i++)if (d[i]==0){S=i;break;} Time=0; Root[0]=0;cnt=0; dfs(S); LL Ans=0; for (int i=1;i<=n;i++){ int bj=GET(K/b[a[i]]); Ans+=Find(Root[ed[i]],1,m,1,bj)-Find(Root[st[i]],1,m,1,bj); } cout<<Ans<<endl;}int main(){ //freopen("1.txt","r",stdin); int Case;scanf("%d",&Case); while (Case--)work(); return 0;}
题解二:树状数组
dfs时,每遇到一个点,把值插入树状数组;每离开一个点,把值从树状数组中移除。
对于当前搜到的点i,统计树状数组中<=k/a[i]的个数,(i点是作为子孙节点统计数量)
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