题目链接
容易发现(a,b,c)肯定是在一条直链上的。
定义(size(u))表示以(u)为根的子树大小（不包括(u)）
分两种情况，
1、(b)是(a)的祖先，对答案的贡献是
[min(deep(p)-1,k)*size(p)
]
显然是可以直接算的。
2、(b)是(a)的孩子，对答案的贡献为
[sum_{ ext{v是u的孩子且deep(v)<=deep(u)+k}}size[v]
]
后半段就可以主席树来维护了。
下标为深度，值为(sum size)，因为同一子树内(dfs)序是连续的，就像树剖那样用主席树维护就好了。
(为什么我总是不记得开(long long)?)
#include <cstdio>
typedef long long ll;
int s; char ch;
inline int read(){
s = 0; ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
return s;
}
inline void Chkmax(int &a, int b){
if(b > a) a = b;
}
inline int min(int a, int b){
return a > b ? b : a;
}
const int MAXN = 300010;
const int MAXMLOGN = 60000010;
int num, maxdeep, cnt, ID, n, m, a, b;
int head[MAXN], deep[MAXN], dfn[MAXN], pos[MAXN], root[MAXN], size[MAXN];
struct Tree{
ll val;
int lc, rc;
}t[MAXMLOGN];
inline void pushup(int now){
t[now].val = t[t[now].lc].val + t[t[now].rc].val;
}
int build(int l, int r){
int id = ++cnt;
if(l == r) return id;
int mid = (l + r) >> 1;
t[id].lc = build(l, mid);
t[id].rc = build(mid + 1, r);
return id;
}
int insert(int now, int l, int r, int x, int y){
int id = ++cnt; t[id] = t[now];
if(l == r){ t[id].val += y; return id; }
int mid = (l + r) >> 1;
if(x <= mid) t[id].lc = insert(t[now].lc, l, mid, x, y);
else t[id].rc = insert(t[now].rc, mid + 1, r, x, y);
pushup(id);
return id;
}
ll query(int p, int q, int l, int r, int wl, int wr){
if(r < wl || l > wr) return 0;
if(l >= wl && r <= wr) return t[q].val - t[p].val;
int mid = (l + r) >> 1; ll ans = 0;
ans += query(t[p].lc, t[q].lc, l, mid, wl, wr);
ans += query(t[p].rc, t[q].rc, mid + 1, r, wl, wr);
return ans;
}
struct Edge{
int next, to;
}e[MAXN << 1];
inline void Add(int from, int to){
e[++num].to = to; e[num].next = head[from]; head[from] = num;
e[++num].to = from; e[num].next = head[to]; head[to] = num;
}
int dfs(int u, int fa){
Chkmax(maxdeep, deep[u] = deep[fa] + 1);
dfn[u] = ++ID;
for(int i = head[u]; i; i = e[i].next)
if(e[i].to != fa)
size[u] += dfs(e[i].to, u);
return size[u] + 1;
}
int main(){
n = read(); m = read();
for(int i = 1; i < n; ++i)
Add(read(), read());
dfs(1, 0);
for(int i = 1; i <= n; ++i)
pos[dfn[i]] = i;
root[0] = build(1, maxdeep);
for(int i = 1; i <= n; ++i)
root[i] = insert(root[i - 1], 1, maxdeep, deep[pos[i]], size[pos[i]]);
for(int i = 1; i <= m; ++i){
a = read(); b = read();
printf("%lld
", (ll)min(deep[a] - 1, b) * size[a] + query(root[dfn[a] - 1], root[dfn[a] + size[a]], 1, maxdeep, deep[a] + 1, deep[a] + b));
}
return 0;
}