首先嘛，还是太弱了，想了好久QAQ
然后，这道题么，明显就是求sigma（size[x]） (x是y的儿子且层树小于k) 然后就可以发现：把前n个节点按深度建可持久化线段树，就能用前缀和维护了
其实不难打= =
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 300010
#define maxm 30000100
struct edges{
int to,next;
}edge[maxn*2];
int next[maxn],l,num;
int addedge(int x,int y){
edge[++l]=(edges){x,next[y]};next[y]=l;
edge[++l]=(edges){y,next[x]};next[x]=l;
return 0;
}
int id[maxn],d[maxn],s[maxn],e[maxn],b[maxn];
int dfs(int x,int y){
id[++num]=x;
d[x]=d[y]+1;
b[x]=num;
for (int i=next[x];i;i=edge[i].next){
if (edge[i].to!=y) {
dfs(edge[i].to,x);
s[x]+=s[edge[i].to]+1;
}
}
e[x]=num;
return 0;
}
struct node{
int lc,rc;long long s;
}t[maxm];
#define lc(x) t[x].lc
#define rc(x) t[x].rc
#define s(x) t[x].s
#define mid ((l+r)>>1)
int build(int x,int l,int r){
if (l!=r) {
lc(x)=build(++num,l,mid);
rc(x)=build(++num,mid+1,r);
}
return x;
}
int ins(int x,int l,int r,int c,int z){
int y=++num;
if (l==r) {s(y)=s(x)+z;return y;}
lc(y)=lc(x);rc(y)=rc(x);
if (mid<c) rc(y)=ins(rc(x),mid+1,r,c,z);
else lc(y)=ins(lc(x),l,mid,c,z);
s(y)=s(lc(y))+s(rc(y));
return y;
}
long long sum(int x,int l,int r,int x1,int y1){
if (l>y1||r<x1) return 0;
if (x1<=l&&r<=y1) return s(x);
if (l==r) return s(x);
return sum(lc(x),l,mid,x1,y1)+sum(rc(x),mid+1,r,x1,y1);
}
int root[maxn];
int main(){
int n,q;
scanf("%d%d",&n,&q);
for (int i=1;i<n;i++) {
int x,y;
scanf("%d%d",&x,&y);
addedge(x,y);
}
dfs(1,0);
num=0;l=0;
root[0]=++num;
build(1,1,n);
for (int i=1;i<=n;i++)root[i]=ins(root[i-1],1,n,d[id[i]],s[id[i]]);
for (int i=1;i<=q;i++) {
int u,v;
scanf("%d%d",&u,&v);
printf("%lld ",s[u]*1ll*min(v,d[u]-1)+sum(root[e[u]],1,n,d[u]+1,d[u]+v)-sum(root[b[u]],1,n,d[u]+1,d[u]+v));
}
return 0;
}