link

G.小沙的身法

由于给出的是一个树，所以两点间简单路径唯一。考虑极端情况，给出的n个点构成一条链，可以用前缀和求解，所以容易想到用lca通过类似方法计算。

const int maxn = 1e6 + 10;
int n, m;
int vis2[maxn];
int a[maxn];
int Log2[maxn], fa[maxn][30], dep[maxn];
int head[maxn];
bool vis[maxn];
long long zheng[maxn], fan[maxn];
struct Edge {
    int to, dis = 1, next;
}edge[maxn << 1];
int p;
void add_edge(int u, int v) {
    p++;
    edge[p].to = v;
    edge[p].next = head[u];
    head[u] = p;
}
void dfs(int cur, int fath = 0) {
    if(vis[cur]) return;
    vis[cur] = true;
    zheng[cur] = zheng[fath] + max(0, a[cur] - a[fath]);
    fan[cur] = fan[fath] + max(0, a[fath]-a[cur]);
    dep[cur] = dep[fath] + 1;
    fa[cur][0] = fath;
    for(int i = 1; i <= Log2[dep[cur]]; i++)
        fa[cur][i] = fa[fa[cur][i-1]][i-1];
    for(int i = head[cur]; i; i = edge[i].next)
        dfs(edge[i].to, cur);
}
void init() {
    for(int i = 1; i <= n; i++) {
        dep[i] = 0;
        head[i] = 0;
    }
    p = 0;
    for(int i = 2; i <= n; i++)
        Log2[i] = Log2[i / 2] + 1;
}
int lca(int a, int b) {
    if(dep[a] > dep[b])
        swap(a, b);
    while(dep[a] != dep[b])
        b = fa[b][Log2[dep[b]-dep[a]]];
    if(a == b)
        return a;
    for(int k = Log2[dep[a]]; k >= 0; k--)  //跳跃长度从长到短
        if(fa[a][k] != fa[b][k]) {
            a = fa[a][k];
            b = fa[b][k];
        }
    return fa[a][0];
}
void solve() {
    cin >> n >> m;
    init();
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        vis2[u]++; vis2[v]++;
        add_edge(u, v);
        add_edge(v, u);
    }
    for(int i = 1; i<= n; i++) {
        if(vis2[i] == 1) {
            dfs(i);
            break;
        }
    }
    while(m--) {
        int u, v;
        cin >> u >> v;
        if(u == v) {
            cout << a[u] << endl;
            continue;
        }
        int l = lca(u, v);
        ll ans = zheng[v] + fan[u] - fan[l] - zheng[l] + a[u];
        cout << ans << endl;
    }
}