Jamie and Tree
Jamie and Tree
https://ac.nowcoder.com/acm/problem/112538
Jamie and Tree
操作一
- 直接更新;
操作二
第一步先找,这时有一个结论中最深的,然后分情况讨论。
不在的子树上:
直接区间更新
在的子树上:
先把整棵树更新一遍+x,然后找到路径上与的儿子节点,然后更新他的子树-x
操作三:
不在的子树上:
直接
在的子树上:
类似操作二。
最后,操作二要特判一下,操作三要特判一下,这个时候直接修改或者查询整个的区间。
#include <bits/stdc++.h> #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1 using namespace std; typedef long long ll; const int N = 1e5 + 10; int head[N], to[N << 1], nex[N << 1], cnt = 1, root; int son[N], sz[N], dep[N], fa[N], top[N], rk[N], id[N], l[N], r[N], value[N], n, m, tot; ll sum[N << 2], lazy[N << 2]; void add(int x, int y) { to[cnt] = y; nex[cnt] = head[x]; head[x] = cnt++; } void dfs1(int rt, int f) { dep[rt] = dep[f] + 1; sz[rt] = 1, fa[rt] = f; for (int i = head[rt]; i; i = nex[i]) { if (to[i] == f) { continue; } dfs1(to[i], rt); sz[rt] += sz[to[i]]; if (!son[rt] || sz[to[i]] > sz[son[rt]]) { son[rt] = to[i]; } } } void dfs2(int rt, int tp) { rk[++tot] = rt, id[rt] = tot; top[rt] = tp; l[rt] = r[rt] = tot; if (!son[rt]) { return ; } dfs2(son[rt], tp); for (int i = head[rt]; i; i = nex[i]) { if (to[i] == fa[rt] || to[i] == son[rt]) { continue; } dfs2(to[i], to[i]); } r[rt] = tot; } void push_down(int rt, int l, int r) { if (lazy[rt]) { lazy[ls] += lazy[rt], lazy[rs] += lazy[rt]; sum[ls] += 1ll * (mid - l + 1) * lazy[rt]; sum[rs] += 1ll * (r - mid) * lazy[rt]; lazy[rt] = 0; } } void push_up(int rt) { sum[rt] = sum[ls] + sum[rs]; } void build(int rt, int l, int r) { if (l == r) { sum[rt] = value[rk[l]]; return ; } build(lson); build(rson); push_up(rt); } void update(int rt, int l, int r, int L, int R, int w) { if (l >= L && r <= R) { lazy[rt] += w; sum[rt] += 1ll * (r - l + 1) * w; return ; } push_down(rt, l, r); if (L <= mid) update(lson, L, R, w); if (R > mid) update(rson, L, R, w); push_up(rt); } ll query(int rt, int l, int r, int L, int R) { if (l >= L && r <= R) return sum[rt]; push_down(rt, l, r); ll ans = 0; if (L <= mid) ans += query(lson, L, R); if (R > mid) ans += query(rson, L, R); return ans; } 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; } int Max(int x, int y) { return dep[x] > dep[y] ? x : y; } void update(int x, int y, int value) { while (top[x] != top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); update(1, 1, n, id[x], id[top[x]], value); x = fa[top[x]]; } if (dep[x] > dep[y]) swap(x, y); update(1, 1, n, id[x], id[y], value); } ll query(int x, int y) { ll ans = 0; while (top[x] != top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); ans += query(1, 1, n, id[x], id[top[x]]); x = fa[top[x]]; } if (dep[x] > dep[y]) swap(x, y); ans += query(1, 1, n, id[x], id[y]); return ans; } int get(int u) { int v = root; while (top[v] != top[u]) { if (fa[top[v]] == u) return top[v]; v = fa[top[v]]; } return son[u]; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); scanf("%d %d", &n, &m); for (int i = 1; i <= n; i++) { scanf("%d", &value[i]); } for (int i = 1; i < n; i++) { int x, y; scanf("%d %d", &x, &y); add(x, y); add(y, x); } dfs1(1, 0); dfs2(1, 1); build(1, 1, n); root = 1; for (int i = 1; i <= m; i++) { int op; scanf("%d", &op); if (op == 1) { scanf("%d", &root); } else if (op == 2) { int u, v, x; scanf("%d %d %d", &u, &v, &x); int lca = Max(Max(Lca(u, v), Lca(root, v)), Lca(root, u)); if (lca == root) { update(1, 1, n, 1, n, x); } else { if (id[root] < l[lca] || id[root] > r[lca]) { update(1, 1, n, l[lca], r[lca], x); } else { lca = get(lca); update(1, 1, n, 1, n, x); update(1, 1, n, l[lca], r[lca], -x); } } } else { int v; scanf("%d", &v); if (v == root) { printf("%lld\n", query(1, 1, n, 1, n)); } else { if (id[root] < l[v] || id[root] > r[v]) { printf("%lld\n", query(1, 1, n, l[v], r[v])); } else { ll ans = query(1, 1, n, 1, n); v = get(v); ans -= query(1, 1, n, l[v], r[v]); printf("%lld\n", ans); } } } } return 0; }