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)

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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

12 31 21 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\n",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\n",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\n",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\n",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\n",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\n",ans);    }    return 0;}

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