題意

題目連結

Sol

一步一步的來考慮

\(25 \%\)：直接\(O(nm)\)的暴力

鏈的情況：維護兩個差分陣列，分別表示從左向右和從右向左的貢獻，

\(S_i = 1\)：統計每個點的子樹內有多少起點即可

\(T_i = 1\)：同樣還是差分的思想，由於每個點 能對其產生的點的深度是相同的(假設為\(x\))，那麼訪問該點時記錄下\(dep[x]\)的數量，將結束時\(dep[x]\)的數量與其做差即可

滿分做法和上面類似，我們考慮把每個點的貢獻都轉換到子樹內統計

對於每次詢問，拆為\(S->lca, lca -> T\)兩種(從下到上 / 從上到下)

從上往下需要滿足的條件：\(dep[i] - w[i] = dep[T] - len\)

從下往上需要滿足的條件：\(dep[i] + w[i] = dep[s]\)

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, B = 20;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, ans[MAXN], dep[MAXN], top[MAXN], son[MAXN], siz[MAXN], fa[MAXN], S[MAXN], 
          T[MAXN], w[MAXN], tmp[MAXN], num2[MAXN], sum1[MAXN], sum2[MAXN], Lca[MAXN];
int *num1;//上 -> 下 
vector<int> up[MAXN], da[MAXN], dc[MAXN];
vector<int> v[MAXN];
void dfs(int x, int _fa) {
    dep[x] = dep[_fa] + 1; siz[x] = 1; fa[x] = _fa;
    for(int i = 0, to; i < v[x].size(); i++) {
        if((to = v[x][i]) == _fa) continue;
        dfs(to, x);
        siz[x] += siz[to];
        if(siz[to] > siz[son[x]]) son[x] = to;
    }
}
void dfs2(int x, int topf) {
    top[x] = topf;
    if(!son[x]) return ;
    dfs2(son[x], topf);
    for(int i = 0, to; i < v[x].size(); i++) 
        if(!top[to = v[x][i]]) dfs2(to, to);
}
int LCA(int x, int y) {
    while(top[x] ^ top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        x = fa[top[x]];
    }
    return dep[x] < dep[y] ? x : y;
}
void Deal(int s, int t, int id) {// from s to t
    int lca = LCA(s, t); Lca[id] = lca;
    up[lca].push_back(s);//from down to up
    int dis = dep[s] + dep[t] - 2 * dep[lca];
    sum2[s]++;
    da[t].push_back(dep[t] - dis);//increase
    dc[lca].push_back(dep[t] - dis);//decrase
}
void Find(int x) {
    int t1 = num1[dep[x] - w[x]], t2 = num2[dep[x] + w[x]];// 1: 從上往下   2：從下往上 
    for(int i = 0, to; i < v[x].size(); i++) {
        if((to = v[x][i]) == fa[x]) continue;
        Find(to);
    }
    num2[dep[x]] += sum2[x];
    for(int i = 0; i < da[x].size(); i++) num1[da[x][i]]++;
    ans[x] += num2[dep[x] + w[x]] - t2 + num1[dep[x] - w[x]] - t1;
    for(int i = 0; i < up[x].size(); i++) num2[dep[up[x][i]]]--;
    for(int i = 0; i < dc[x].size(); i++) num1[dc[x][i]]--;
}
int main() {
    //freopen("a.in", "r", stdin); freopen("a.out", "w", stdout);
    num1 = tmp + (int)3e5 + 10;
    N = read(); M = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }
    dep[0] = -1; dfs(1, 0); dfs2(1, 1);
    //for(int i = 1; i <= N; i++, puts("")) for(int j = 1; j <= N; j++) printf("%d %d %d\n", i, j, LCA(i, j));  
    for(int i = 1; i <= N; i++) w[i] = read();
    for(int i = 1; i <= M; i++) S[i] = read(), T[i] = read(), Deal(S[i], T[i], i);
    Find(1);
    for(int i = 1; i <= M; i++) if(dep[S[i]] - dep[Lca[i]] == w[Lca[i]]) ans[Lca[i]]--;
    for(int i = 1; i <= N; i++) printf("%d ", ans[i]);
    return 0;
}