BZOJ3653: Laughing

Description

Set T to a tree with a root tree, we do the following definitions:
• Set A and B as two different nodes in T. If a is an ancestor of B, then it is called "a than B does not know
Where did gaoming go? "
• Set A and B as two different nodes in T. If the distance between A and B in the tree does not exceed a given
Constant X, then called "A and B" laughing.
Given a root tree t of n nodes, the nodes are numbered 1 to n and the root node is node # 1th. You need
To answer Q queries, ask for a given two integers p and K, and ask how many ordered triples (A;B;C) satisfy:
1. A, B and C are three different points in T, and A is the P node;
2. A and b are better than C to know where the wise go;
3. A and B are laughing. The constant in the laughing here is the given K.

Input

The first line of the input file contains two positive integers n and q, each representing the number of points and queries that have a root tree. Next n-1 lines, each line describes an edge on a tree. Each row contains two integers u and V, which represents an edge between the node U and v.
Next the Q line, each line describes an operation. Line I contains two integers representing the p and K of the I inquiry respectively.

Output

Output Q line, each line corresponding to a query, representative of the answer to the question.

Sample Input5 3
1 2
1 3
2 4
4 5
2 2
4 1
2 3
Sample Output
3
1
3
HINT

1<=p<=n

1<=k<=n

n<=300000

q<=300000

Ah, the clam problem ... According to the definition is not difficult to find that ABC three points should be the same in a straight chain, then we press B on the top and the following 2 categories the first case is very simple, because a subtree of all nodes except a can be used as C. The second case is a partial-order structure, where B should be within a subtree and the depth of B does not exceed the depth of a +k. Because the problem can be taken offline, the size information can be maintained by merging it with a tree of balanced trees and line segments. Because the memory is richer, I'm writing a line tree merge.

#include <cstdio> #include <cctype> #include <queue> #include <cstring> #include <algorithm > #define REP (i,s,t) for (int. i=s;i<=t;i++) #define DWN (I,S,T) for (int. i=s;i>=t;i--) #define REN for (int i=first[x ];i;i=next[i]) using namespace Std;const int Buffersize=1<<16;char Buffer[buffersize],*head,*tail;inline Char Getchar () {if (head==tail) {int l=fread (Buffer,1,buffersize,stdin); tail= (Head=buffer) +l;} return *head++;}    inline int read () {int X=0,f=1;char c=getchar (); for (;!    IsDigit (c); C=getchar ()) if (c== '-') f=-1;    for (; IsDigit (c); C=getchar ()) x=x*10+c-' 0 '; return x*f;} typedef long LONG ll;const int Maxn=300010;const int maxnode=maxn*20;struct Query {int k,id,next;} Q[maxn];int n,m,first2[maxn],first[maxn],next[maxn<<1],to[maxn<<1],dep[maxn],siz[maxn],e,cnt;void Addedge (int u,int v) {to[++e]=v;next[e]=first[u];first[u]=e;to[++e]=u;next[e]=first[v];first[v]=e;} void addquery (int x,int k,int ID) {q[++cnt]= (Query) {k,id,first2[x]};first2[x]=cnt;}ll s[maxnode],ans[maxn];int ls[maxnode],rs[maxnode],root[maxn],tot;void Build (int& o,int l,int r,int p,int val) { if (!o) o=++tot;s[o]+=val;if (l==r) return;int mid=l+r>>1;if (p<=mid) build (ls[o],l,mid,p,val); else Build (rs[ O],mid+1,r,p,val);} int merge (int x,int y) {if (!x) return y;if (!y) return X;s[x]+=s[y];ls[x]=merge (Ls[x],ls[y]); Rs[x]=merge (Rs[x],rs[y]); return x;} ll query (int o,int l,int r,int x) {if (!o) return 0;if (l==r) return s[o];int mid=l+r>>1;if (x<=mid) return query (LS [O],l,mid,x]; return S[ls[o]]+query (rs[o],mid+1,r,x);} void Dfs (int x,int fa) {Siz[x]=1;dep[x]=dep[fa]+1;ren if (TO[I]!=FA) DFS (to[i],x), Siz[x]+=siz[to[i]],root[x]=merge ( Root[x],root[to[i]]); for (int i=first2[x];i;i=q[i].next) {int k=q[i].k,j=q[i].id;ans[j]= (LL) (siz[x]-1) *min (k,dep[x ]-1); Ans[j]+=query (Root[x],1,n,min (N,dep[x]+k));} Build (root[x],1,n,dep[x],siz[x]-1);} int main () {n=read (); M=read (); Rep (i,2,n) Addedge (read (), read ()), Rep (i,1,m) {int p=read (), K=read (); AddQuery (p,k,i);} DFS (1,0); Rep (i,1,m) printf ("%lld\n", Ans[i]); return 0;} 

　　

BZOJ3653: Laughing