2016 ACM/ICPC Dalian Online-1010 Weak Pair
2016-09-14 10:34
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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
;
int st
,ed
,Time;
int a
,b
,d
,Root
;
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)
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
;
int st
,ed
,Time;
int a
,b
,d
,Root
;
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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