HDU 5877 2016大连网络赛 Weak Pair(树状数组，线段树，动态开点，启发式合并，可持久化线段树)

Weak Pair

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1468    Accepted Submission(s): 472

Problem Description

You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of nodes (u,v) is said to be weak if
  (1) u is an ancestor of v (Note: In this problem a node u is not considered an ancestor of itself);
  (2) au×av≤k.

Can you find the number of weak pairs in the tree?

 

Input

There are multiple cases in the data set.
  The first line of input contains an integer T denoting number of test cases.
  For each case, the first line contains two space-separated integers, N and k, respectively.
  The second line contains N space-separated integers, denoting a1 to aN.
  Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes u and v , where node u is the parent of node v.

  Constrains: 
  
  1≤N≤105 
  
  0≤ai≤109 
  
  0≤k≤1018

 

Output

For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

 

Sample Input

1
2 3
1 2
1 2

 

Sample Output

1

这是一道很好的数据结构的题目：
可以用很多方法写
首先思路是：dfs这颗树，每到一个节点，都计算这个节点的祖先中满足条件的有几个
而计算这个就需要维护一个序列，并且高效的得出多少个祖先满足条件。
即在序列中找到小于k/a[i]的数有多少个，很容易想到用树状数组和线段树。
权值1e9需要离散化。
树状数组：
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <map>

using namespace std;
const int maxn=1e5;
typedef long long int LL;
int n;
LL k;
LL a[maxn+5];
struct Node
{
    int value;
    int next;
}edge[maxn*2+5];
int head[maxn+5];
int vis[maxn+5];
int tot;
int c[maxn*2+5];
LL b[maxn+5];
LL e[maxn*2+5];
map<LL,int> m;
void add(int x,int y)
{
    edge[tot].value=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int num)
{
    while(x<=n*2)
    {
        c[x]+=num;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int _sum=0;
    while(x>0)
    {
        _sum+=c[x];
        x-=lowbit(x);
    }
    return  _sum;
}
LL ans;
void dfs(int root)
{
    vis[root]=1;
    for(int i=head[root];i!=-1;i=edge[i].next)
    {
        int v=edge[i].value;
        if(!vis[v])
        {
            ans+=sum(m[b[v]]);
            update(m[a[v]],1);
            dfs(v);
            update(m[a[v]],-1);
        }
    }
}

void init()
{
    memset(c,0,sizeof(c));
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    tot=0;
}
int tag[maxn+5];
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        init();
        int cnt=n;
        m.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            e[i]=a[i];
            if(a[i]==0)
                m[a[i]]=2*n;
            else
            {
                b[i]=k/a[i];
                e[++cnt]=b[i];
            }
        }
        sort(e+1,e+cnt+1);
        int tot=1;
        for(int i=1;i<=cnt;i++)
        {
            if(!m.count(e[i]))
                m[e[i]]=tot++;
        }
        memset(tag,0,sizeof(tag));
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y);
            tag[y]++;
        }
        int root;
        for(int i=1;i<=n;i++)
        {
            if(tag[i]==0)
                root=i;
        }
        ans=0;
        update(m[a[root]],1);
        dfs(root);
        printf("%lld
",ans);
    }
    return 0;
}

线段树：
<pre name="code" class="html">#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
#include <string>
#include <map>
#include <vector>

using namespace std;
typedef long long int LL;
const int maxn=1e5;
vector<int> v[maxn+5];
int sum[maxn*8+5];
int n;
LL k;
LL a[maxn+5];
LL b[maxn+5];
LL e[maxn*2+5];
int aa[maxn+5];
int bb[maxn+6];
map<LL,int> m;

void PushUp(int node)
{
    sum[node]=sum[node<<1]+sum[node<<1|1];
}
void update(int node,int begin,int end,int ind,int num)
{
    if(begin==end)
    {
        sum[node]+=num*(end-begin+1);
        return;
    }
    int m=(begin+end)>>1;
    if(ind<=m)
        update(node<<1,begin,m,ind,num);
    else
        update(node<<1|1,m+1,end,ind,num);
    PushUp(node);
}
LL Query(int node,int begin,int end,int left,int right)
{
    if(left<=begin&&end<=right)
        return sum[node];
    int m=(begin+end)>>1;
    LL ret=0;
    if(left<=m)
        ret+=Query(node<<1,begin,m,left,right);
    if(right>m)
        ret+=Query(node<<1|1,m+1,end,left,right);
    PushUp(node);
    return ret;
}
int tag[maxn+5];
LL ans;
void dfs(int root)
{
    int len=v[root].size();
    for(int i=0;i<len;i++)
    {
        int w=v[root][i];
        ans+=Query(1,1,2*n,1,bb[w]);
        update(1,1,2*n,aa[w],1);
        dfs(v[root][i]);
        update(1,1,2*n,aa[w],-1);
    }
}
void init()
{

    memset(sum,0,sizeof(sum));
    memset(tag,0,sizeof(tag));
}
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        int cnt=0;
        init();
        m.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            e[++cnt]=a[i];
            b[i]=k/a[i];
            e[++cnt]=b[i];
            v[i].clear();
        }
        sort(e+1,e+cnt+1);
        int cot=1;
        for(int i=1;i<=cnt;i++)
        {
            if(!m.count(e[i]))
                m[e[i]]=cot++;
        }
        for(int i=1;i<=n;i++)
        {
            aa[i]=m[a[i]];
            bb[i]=m[b[i]];
        }
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            v[x].push_back(y);
            tag[y]++;
        }
        int root;
        for(int i=1;i<=n;i++)
        {
            if(tag[i]==0)
                root=i;
        }
        ans=0;
        update(1,1,2*n,m[a[root]],1);
        dfs(root);
        printf("%lld
",ans);
    }
    return 0;
}
其实用线段树也可以不离散的方法做，是线段树的动态开点，动态开点就是用到了这个点再去开，不用的点不用开
这样在0到1e18的范围内，最多储存的点也就n个叶子节点，开个8*n的空间就足够了

</pre><pre code_snippet_id="1877993" snippet_file_name="blog_20160912_2_1715825" name="code" class="html"><pre name="code" class="html">#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <string>
#include <stdio.h>
#include <vector>

using namespace std;
const int maxn=1e5;
const long long int len=1e18;
typedef long long int LL;
LL a[maxn+5];
LL b[maxn+5];
int n;
LL k;
vector<int> v[maxn+5];
struct Node
{
    int lch,rch;
    LL sum;
    Node(){};
    Node(int lch,int rch,LL sum)
    {
        this->lch=lch;
        this->rch=rch;
        this->sum=sum;
    }
}tr[maxn*100+5];
int p;
void PushUp(int node)
{
    tr[node].sum=tr[tr[node].lch].sum+tr[tr[node].rch].sum;
}

int newnode()
{
    tr[++p]=Node(-1,-1,0);
    return p;
}
void update(int node,LL begin,LL end,LL ind,int num)
{
    if(begin==end)
    {
        tr[node].sum+=num;
        return;
    }
    LL m=(begin+end)>>1;
    if(tr[node].lch==-1) tr[node].lch=newnode();
    if(tr[node].rch==-1) tr[node].rch=newnode();
    if(ind<=m)
        update(tr[node].lch,begin,m,ind,num);
    else
        update(tr[node].rch,m+1,end,ind,num);
    PushUp(node);
}
LL query(int node,LL begin,LL end,LL left,LL right)
{
    if(node==-1)
        return 0;
    if(left<=begin&&end<=right)
        return tr[node].sum;
    LL m=(begin+end)>>1;
    LL ret=0;
    if(left<=m)
        ret+=query(tr[node].lch,begin,m,left,right);
    if(right>m)
        ret+=query(tr[node].rch,m+1,end,left,right);
    PushUp(node);
    return ret;

}
int tag[maxn+5];
LL ans;
void dfs(int root)
{
    int len1=v[root].size();
    for(int i=0;i<len1;i++)
    {
        int w=v[root][i];
        ans+=query(1,0,len,0,b[w]);
        update(1,0,len,a[w],1);
        dfs(w);
        update(1,0,len,a[w],-1);
    }
}
void init()
{
    memset(tag,0,sizeof(tag));
    p=0;
    newnode();
}
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            b[i]=k/a[i];
            v[i].clear();
        }
        init();
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            v[x].push_back(y);
            tag[y]++;
        }
        int root;
        for(int i=1;i<=n;i++)
        {
            if(!tag[i])
                root=i;
        }
        ans=0;
        update(1,0,len,a[root],1);
        dfs(root);
        printf("%lld
",ans);
    }
    return 0;
}

还可以自下而上，用线段树的启发合并，计算每一个节点的所有子节点对他的贡献

关于线段树的启发式合并，有必要再写一篇博客总结一下

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
const int maxn=1e5;
typedef long long int LL;
int rt[maxn*100+5];
int ls[maxn*100+5];
int rs[maxn*100+5];
LL sum[maxn*100+5];
int a[maxn+5];
LL k;
int n;
int p;
int l,r;
int newnode()
{
    sum[p]=ls[p]=rs[p]=0;
    return p++;
}
void build(int &node,int begin,int end,LL val)
{
    if(!node) node=newnode();
    sum[node]=1;
    if(begin==end) return;
    LL mid=(begin+end)>>1;
    if(val<=mid) build(ls[node],begin,mid,val);
    else build(rs[node],mid+1,end,val);
}
LL Query(int node,int begin,int end,LL val)
{
    if(!node||val<begin) return 0;
    if(begin==end) return sum[node];
    LL mid=(begin+end)>>1;
    if(val<=mid) return Query(ls[node],begin,mid,val);
    else return sum[ls[node]]+Query(rs[node],mid+1,end,val);
}
void mergge(int &x,int y, int begin,int end)
{
    if(!x||!y) {x=x^y;return;}
    sum[x]+=sum[y];
    if(begin==end) return;
    LL mid=(begin+end)>>1;
    mergge(ls[x],ls[y],begin,mid);
    mergge(rs[x],rs[y],mid+1,end);
}
struct Node
{
    int value;
    int next;
}edge[maxn*2+5];
int head[maxn+5];
int tot;
void add(int x,int y)
{
    edge[tot].value=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
LL ans;
void dfs(int root)
{
    for(int i=head[root];i!=-1;i=edge[i].next)
    {
        int w=edge[i].value;
           dfs(w);
           mergge(rt[root],rt[w],l,r);
    }
    ans+=Query(rt[root],l,r,k/a[root]);
    if(k>=1ll*a[root]*a[root])
        ans--;
}
int tag[maxn+5];
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        p=1;
        memset(tag,0,sizeof(tag));
        memset(head,-1,sizeof(head));

        tot=0;
        l=1e9;r=0;
        for(int i=1;i<=n;i++)
        {
             scanf("%d",&a[i]);
             l=min(l,a[i]);r=max(r,a[i]);
        }
        for(int i=1;i<=n;i++)
            build(rt[i]=0,l,r,a[i]);


        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y);
            tag[y]++;
        }
        int root;
        for(int i=1;i<=n;i++)
           if(tag[i]==0) root=i;
        ans=0;
        dfs(root);
        printf("%lld
",ans);
    }
    return 0;
}

也可以用拓扑排序，自下而上进行启发式合并，

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <queue>

using namespace std;
const int maxn=1e5;
typedef long long int LL;
int rt[maxn*100+5];
int ls[maxn*100+5];
int rs[maxn*100+5];
LL sum[maxn*100+5];
int a[maxn+5];
int f[maxn+5];
LL k;
int n;
int p;
int l,r;
queue<int> q;
int newnode()
{
    sum[p]=ls[p]=rs[p]=0;
    return p++;
}
void build(int &node,int begin,int end,LL val)
{
    if(!node) node=newnode();
    sum[node]=1;
    if(begin==end) return;
    LL mid=(begin+end)>>1;
    if(val<=mid) build(ls[node],begin,mid,val);
    else build(rs[node],mid+1,end,val);
}
LL Query(int node,int begin,int end,LL val)
{
    if(!node||val<begin) return 0;
    if(begin==end) return sum[node];
    LL mid=(begin+end)>>1;
    if(val<=mid) return Query(ls[node],begin,mid,val);
    else return sum[ls[node]]+Query(rs[node],mid+1,end,val);
}
void mergge(int &x,int y, int begin,int end)
{
    if(!x||!y) {x=x^y;return;}
    sum[x]+=sum[y];
    if(begin==end) return;
    LL mid=(begin+end)>>1;
    mergge(ls[x],ls[y],begin,mid);
    mergge(rs[x],rs[y],mid+1,end);
}
LL ans;
int tag[maxn+5];
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        p=1;
        memset(tag,0,sizeof(tag));
        l=1e9;r=0;
        for(int i=1;i<=n;i++)
        {
             scanf("%d",&a[i]);
             l=min(l,a[i]);r=max(r,a[i]);
        }
        for(int i=1;i<=n;i++)
            build(rt[i]=0,l,r,a[i]);


        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            tag[x]++;
			f[y]=x;
        }
        for(int i=1;i<=n;i++)
		{
           if(tag[i]==0)
			   q.push(i);
		}
        ans=0;
		while(!q.empty())
		{
			int x=q.front();q.pop();
			if(1LL*a[x]*a[x]<=k) ans--;
			ans+=Query(rt[x],l,r,k/a[x]);
			mergge(rt[f[x]],rt[x],l,r);
			if(!--tag[f[x]]) q.push(f[x]);
			
		}
        printf("%lld
",ans);
    }
    return 0;
}

最后写一种，可持续化线段树的解法。首先将树形转成线形，然后逐个点插入，求一个根节点的子树对根节点的贡献，就是求DFS序列一段区间

小于k/a[i]的有多少个，可持续化线段树利用类似前缀和的原理，tree[r]-tree[l-1]就是l到r这一段区间所有点的线段树

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <stack>

using namespace std;
const int maxn=1e5;
typedef long long int LL;
int rt[maxn*100+5];
int ls[maxn*100+5];
int rs[maxn*100+5];
LL sum[maxn*100+5];
int p;
int n;
LL k;
int l,r;
void update(int &node,int l,int r,int val)
{
	
    ls[p]=ls[node];rs[p]=rs[node];
    sum[p]=sum[node];node=p;
	p++;
    
    if(l==r)
    {
        sum[node]++;
        return;
    }
    sum[node]++;
    int mid=(l+r)>>1;
    if(val<=mid) update(ls[node],l,mid,val);
    else update(rs[node],mid+1,r,val);
}
LL query(int node,int l,int r,LL val)
{
	if(val<l) return 0;
    if(!node) return 0;
    if(l==r) return sum[node];
    LL mid=(l+r)>>1;
    if(val<=mid) return query(ls[node],l,mid,val);
    else return sum[ls[node]]+query(rs[node],mid+1,r,val);
}
struct Node
{
    int value;
    int next;
}edge[maxn*2+5];
int head[maxn+5];
int tot;
void add(int x,int y)
{
    edge[tot].value=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
int res[maxn*2];
int a[maxn+5];
int cot;
void dfs(int root)
{
    res[cot++]=root;
    for(int i=head[root];i!=-1;i=edge[i].next)
    {
        int w=edge[i].value;
        dfs(w);
    }
    res[cot++]=root;
}
int tag[maxn+5];
int flag[maxn+5];
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    while(t--)
    {
        scanf("%d%lld",&n,&k);
        l=1e9;r=0;
        for(int i=1;i<=n;i++)
        {
             scanf("%d",&a[i]);
             l=min(l,a[i]);r=max(r,a[i]);
        }
           
        memset(head,-1,sizeof(head));
        memset(tag,0,sizeof(tag));
        tot=0;
        p=1;
        for(int i=1;i<=n-1;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y);
            tag[y]++;
        }
        int root;
        for(int i=1;i<=n;i++)
        {
            if(!tag[i])
                root=i;
        }
        cot=0;
        dfs(root);
        memset(flag,0,sizeof(flag));
        update(rt[res[0]],l,r,a[res[0]]);
        flag[res[0]]=1;
        LL ans=0;
		int now=0;
        for(int i=1;i<cot;i++)
        {
            if(flag[res[i]]==1)
            {
				LL ans1=query(rt[res[now]],l,r,k/a[res[i]]);
				LL ans2=query(rt[res[i]],l,r,k/a[res[i]]);
				//cout<<ans1<<" "<<ans2<<endl;
                ans+=ans1-ans2;
                continue;
            }
            flag[res[i]]=1;
            update(rt[res[i]]=rt[res[now]],l,r,a[res[i]]);
			now=i;
        }
        printf("%lld
",ans);
    }
    return 0;
}