南昌网络赛J. Distance on the tree　树链剖分＋主席树

　　Distance on the tree

题目链接

https://nanti.jisuanke.com/t/38229

Describe

DSM(Data Structure Master) once learned about tree when he was preparing for NOIP(National Olympiad in Informatics in Provinces) in Senior High School. So when in Data Structure Class in College, he is always absent-minded about what the teacher says.

The experienced and knowledgeable teacher had known about him even before the first class. However, she didn't wish an informatics genius would destroy himself with idleness. After she knew that he was so interested in ACM(ACM International Collegiate Programming Contest), she finally made a plan to teach him to work hard in class, for knowledge is infinite.

This day, the teacher teaches about trees." A tree with nnn nodes, can be defined as a graph with only one connected component and no cycle. So it has exactly n−1n-1n−1 edges..." DSM is nearly asleep until he is questioned by teacher. " I have known you are called Data Structure Master in Graph Theory, so here is a problem. "" A tree with nnn nodes, which is numbered from 111 to nnn. Edge between each two adjacent vertexes uuu and vvv has a value w, you're asked to answer the number of edge whose value is no more than kkk during the path between uuu and vvv."" If you can't solve the problem during the break, we will call you DaShaMao(Foolish Idiot) later on."

The problem seems quite easy for DSM. However, it can hardly be solved in a break. It's such a disgrace if DSM can't solve the problem. So during the break, he telephones you just for help. Can you save him for his dignity?

Input

In the first line there are two integers n,mn,mn,m, represent the number of vertexes on the tree and queries(2≤n≤10^5,1≤m≤10^5)

The next n−1n-1n−1 lines, each line contains three integers u,v,wu,v,wu,v,w, indicates there is an undirected edge between nodes uuu and vvv with value www. (1≤u,v≤n,1≤w≤10^9)

The next mmm lines, each line contains three integers u,v,ku,v,ku,v,k , be consistent with the problem given by the teacher above. (1≤u,v≤n,0≤k≤10^9)

Output

For each query, just print a single line contains the number of edges which meet the condition.

样例输入1

样例输出1

0
1
2

样例输入2

5 2
1 2 1000000000
1 3 1000000000
2 4 1000000000
3 5 1000000000
2 3 1000000000
4 5 1000000000

样例输出2

题意

给你一棵树，问两个点之间边权小与等于k的数。

题解

智商不够，靠数据结构来凑，弱弱的我直接树剖，然后用主席树，感觉就是把两个板子结合一下。

第一次一遍AC，看了好几遍才确定自己没看错。

代码

 #include<bits/stdc++.h>
 using namespace std;
 #define N 500050
 #define ll long long
 #define INF 123456789
 struct Query{int l,r,tt;}que[N];
 ];
 int totn,root[N];
 ];
 int last[N],tot;
 int n,m,a[N],w[N],num;
 int cnt,fa[N],d[N],size[N],son[N],kth[N],rk[N],top[N];
 void AddEdge(int x,int y,int z)
 {
   edges[++tot]=Edge{x,y,z,last[x]};
   last[x]=tot;
 }
 void dfs1(int u,int pre,int val)
 {
   d[u]=d[pre]+;
   fa[u]=pre;
   size[u]=;
   w[u]=val;
   for(int i=last[u];i;i=edges[i].s)
     {
       Edge &e=edges[i];
       if (e.to==pre)continue;
       dfs1(e.to,u,e.val);
       size[u]+=size[e.to];
       if (size[e.to]>size[son[u]])son[u]=e.to;
     }
 }
 void dfs2(int u,int y)
 {
   rk[u]=++cnt;
   kth[cnt]=u;
   top[u]=y;
   )return;
   dfs2(son[u],y);
   for(int i=last[u];i;i=edges[i].s)
     {
       Edge &e=edges[i];
       if (e.to==son[u]||e.to==fa[u])continue;
       dfs2(e.to,e.to);
     }
 
 }
 void bt(int &x,int l,int r)
 {
     x=++totn;
     if (l==r)return ;
     ;
     bt(tr[x].ls,l,mid);
     bt(tr[x].rs,mid+,r);
 }
 void add(int &x,int last,int l,int r,int p)
 {
     x=++totn;
     tr[x]=tr[last];
     if (l==r){tr[x].sum++;return;}
     ;
     if (p<=mid) add(tr[x].ls,tr[last].ls,l,mid,p);
     ,r,p);
     tr[x].sum=tr[tr[x].ls].sum+tr[tr[x].rs].sum;
 }
 int ask(int ql,int qr,int l,int r,int kk)
 {
   <=l&&r<=kk)return tr[qr].sum-tr[ql].sum;
   ,ans=;
   <=mid)ans+=ask(tr[ql].ls,tr[qr].ls,l,mid,kk);
   ,r,kk);
   return ans;
 }
 int get_sum(int x,int y,int tt)
 {
   ;
   int fx=top[x],fy=top[y];
   while(fx!=fy)
     {
       if (d[fx]<d[fy])swap(x,y),swap(fx,fy);
       ans+=ask(root[rk[fx]-],root[rk[x]],,num,tt);//
       x=fa[fx];fx=top[x];
     }
   if(d[x]<d[y])swap(x,y);
   ans+=ask(root[rk[y]],root[rk[x]],,num,tt);
   return ans;
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
   freopen("aa.in","r",stdin);
 #endif
   ios::sync_with_stdio(false);
   cin>>n>>m;
   ll x,y,z,id,ans;
   ;i<=n-;i++)
     {
       cin>>x>>y>>z;
       a[++num]=z;
       AddEdge(x,y,z);
       AddEdge(y,x,z);
     }
   ;i<=m;i++)
     {
       cin>>que[i].l>>que[i].r>>que[i].tt;
       a[++num]=que[i].tt;
     }
   sort(a+,a+num+);
   num=unique(a+,a+num+)-a-;
   dfs1(,,INF);
   ;i<=n;i++)w[i]=lower_bound(a+,a+num+,w[i])-a;
   dfs2(,);
   bt(root[],,num);
   ;i<=n;i++)
     add(root[i],root[i-],,num,w[kth[i]]);
   ;i<=m;i++)
     {
       que[i].tt=lower_bound(a+,a+num+,que[i].tt)-a;
       int ans=get_sum(que[i].l,que[i].r,que[i].tt);
       printf("%d\n",ans);
     }
 }