通过观察可以发现，当以u为根节点的时候，向他的子节点转移x次，此时差值固定，即等差数列。
我们就可以通过树链剖分，给每一条链上，维护线段树，保存等差数列。
最后再通过求前缀和的方式求得。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 5e5 + 10;
struct node { int to, next; }edge[2*N];
int head[2*N];
int id[N], rk[N], son[N], fa[N], top[N], dep[N], siz[N];
int cnt = 0;
int n, m;
void addedge(int u, int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
void dfs1(int u, int f, int d)
{
fa[u] = f; dep[u] = d; siz[u] = 1;
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == f)continue;
dfs1(v, u, d + 1);
siz[u] += siz[v];
if (son[u] == -1 || siz[son[u]] < siz[v])
{
son[u] = v;
}
}
}
int num = 0;
void dfs2(int u, int tp)
{
top[u] = tp;
id[u] = ++num;
rk[num] = u;
if (son[u] == -1)return;
dfs2(son[u], tp);
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == fa[u]||v==son[u])continue;
dfs2(v, v);
}
}
struct {
ll st;
ll d;
}lazy[N<<2];
void pushdown(int root,int l,int r)
{
int mid = (l + r) >> 1;
lazy[root << 1].d += lazy[root].d;
lazy[root << 1 | 1].d += lazy[root].d;
lazy[root << 1].st += lazy[root].st;
lazy[root << 1 | 1].st += lazy[root].st - 1ll * 2 * (mid - l + 1)*lazy[root].d;
lazy[root].d = lazy[root].st = 0;
}
void update(int l, int r, int root, int lf, int rt, ll st, ll d)
{
if (lf <= l && r <= rt)
{
lazy[root].st += st;
lazy[root].d += d;
return;
}
pushdown(root, l, r);
int mid = (l + r) >> 1;
if (lf <= mid)update(lson, lf, rt, st, d);
if (rt > mid)
{
if (lf <= mid)update(rson, lf, rt, st - 1ll * 2 * (mid - max(lf, l) + 1)*d, d);
else update(rson, lf, rt, st, d);
}
}
int lca(int u, int v)
{
while (top[u] != top[v])
{
if (dep[top[u]] < dep[top[v]])swap(u, v);
u = fa[top[u]];
}
if (dep[u] < dep[v])swap(u, v);
return v;
}
ll ans[N];
void modify(int x, int y)
{
int tp = lca(x, y);
ll depx = dep[x] - dep[tp];
ll depy = dep[y] - dep[tp];
ll tempx = 0;
ll tempy = 0;
while (top[x] != top[tp])
{
tempx = dep[tp] - dep[top[x]] + 1;
update(1, n, 1, id[top[x]], id[x], depx - depy - 1 + 2 * tempx, 1);
x = fa[top[x]];
}
if (x != tp)
update(1, n, 1, id[tp] + 1, id[x], depx - depy - 1, 1);
while (top[y] != top[tp])
{
tempy = dep[tp] - dep[top[y]] + 1;
update(1, n, 1, id[top[y]], id[y], depy - depx - 1 + 2 * tempy, 1);
y = fa[top[y]];
}
if(y!=tp)
update(1, n, 1, id[tp] + 1, id[y], depy - depx - 1, 1);
ans[1] += depx * depy;
}
void rebuild(int l, int r, int root)
{
if (l == r) {
if (l == 1)return;
//cout << "rkl" << rk[l] << endl;
ans[rk[l]] = lazy[root].st;
//cout << ans[rk[l]] << endl;
return;
}
pushdown(root, l, r);
int mid = (l + r) >> 1;
rebuild(lson);
rebuild(rson);
}
void dfs3(int u, int f)
{
for (int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if (v == f)continue;
ans[v] += ans[u];
dfs3(v, u);
}
}
int main()
{
memset(head, -1, sizeof(head));
memset(son, -1, sizeof(son));
n = read(), m = read();
int u, v;
up(i, 0, n - 1)
{
u = read(), v = read();
addedge(u, v);
addedge(v, u);
/*addedge(i + 1, i + 2);
addedge(i + 2, i + 1);*/
}
dfs1(1, 1, 1);
dfs2(1, 1);
upd(i, 0, m - 1)
{
u = read(), v = read();
modify(u, v);
}
rebuild(1, n, 1);
dfs3(1, 0);
upd(i, 1, n)
{
printf("%lld\n", ans[i]);
}
return 0;
}