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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#468278#1163. Another Tree Queries ProblemwcyQwQTL 4ms39976kbC++174.7kb2024-07-08 20:06:242024-07-08 20:06:25

Judging History

你现在查看的是最新测评结果

  • [2024-07-08 20:06:25]
  • 评测
  • 测评结果:TL
  • 用时:4ms
  • 内存:39976kb
  • [2024-07-08 20:06:24]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;
using LL = long long;

const int N = 2e5 + 10;
vector<int> g[N];
int son[N], top[N], sz[N], dep[N], fa[N], dfn[N], rk[N], tim, n;
LL sdp[N], ssz[N];

void dfs1(int u, int p) {
  fa[u] = p, dep[u] = dep[p] + 1, sz[u] = 1;
  for (int v : g[u]) if (v != p) {
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}

void dfs2(int u, int t) {
  dfn[u] = ++tim, rk[tim] = u, top[u] = t;
  if (son[u]) dfs2(son[u], t);
  for (int v : g[u]) if (v != son[u] && v != fa[u]) dfs2(v, v);
}

int find(int u, int v) { // the son of u -> v
  while (top[u] != top[v]) {
    if (fa[top[u]] == v) return top[u];
    u = top[u];
  }
  return son[v];
}

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 u;
}

struct Tag {
  LL x, y, z;
  Tag(LL x = 0, LL y = 0, LL z = 0) : x(x), y(y), z(z) {}
};

struct Node {
  LL sum;
  Tag tg;
  Node(LL sum = 0, Tag tg = Tag()) : sum(sum), tg(tg) {}
};

inline Tag operator+(Tag a, Tag b) { return Tag(a.x + b.x, a.y + b.y, a.z + b.z); }

inline Node operator+(Node a, Node b) { return Node(a.sum + b.sum); }

inline void tag(Node &u, Tag tg, int l, int r) {
  u.sum += (ssz[r] - ssz[l - 1]) * tg.x;
  u.sum += (sdp[r] - sdp[l - 1]) * tg.y;
  u.sum += tg.z * (r - l + 1);
  u.tg = u.tg + tg;
}

struct SGT {
  Node t[N * 4];

  void pushup(int p) { t[p] = t[p * 2] + t[p * 2 + 1]; }

  void pushdown(int p, int l, int r) {
    int m = (l + r) / 2;
    tag(t[p * 2], t[p].tg, l, m);
    tag(t[p * 2 + 1], t[p].tg, m + 1, r);
    t[p].tg = Tag();
  }

  void modify(int p, int l, int r, int ql, int qr, Tag tg) {
    if (ql <= l && r <= qr) return tag(t[p], tg, l, r);
    pushdown(p, l, r);
    int m = (l + r) / 2;
    if (ql <= m) modify(p * 2, l, m, ql, qr, tg);
    if (qr > m) modify(p * 2 + 1, m + 1, r, ql, qr, tg);
    pushup(p);
  }

  void modify(int l, int r, Tag tg) { modify(1, 1, n, l, r, tg); }

  LL query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[p].sum;
    pushdown(p, l, r);
    int m = (l + r) / 2;
    if (qr <= m) return query(p * 2, l, m, ql, qr);
    if (ql > m) return query(p * 2 + 1, m + 1, r, ql, qr);
    return query(p * 2, l, m, ql, qr) + query(p * 2 + 1, m + 1, r, ql, qr);
  }

  LL query(int l, int r) { return query(1, 1, n, l, r); }
} sgt;

void mPath(int u, int v, Tag tg) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    sgt.modify(dfn[top[u]], dfn[u], tg);
    u = fa[top[u]];
  }
  if (dfn[u] > dfn[v]) swap(u, v);
  sgt.modify(dfn[u], dfn[v], tg); 
}

LL qPath(int u) {
  LL ans = 0;
  while (u) {
    ans += sgt.query(dfn[top[u]], dfn[u]);
    u = fa[top[u]];
  }
  return ans;
}

int main() {
  // freopen("A.in", "r", stdin);
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs1(1, 0), dfs2(1, 1);
  for (int i = 1; i <= n; i++) {
    sdp[i] = sdp[i - 1] + dep[rk[i]];
    ssz[i] = ssz[i - 1] + sz[rk[i]];
  }
  LL s1 = 0, s2 = 0;
  auto calc = [&] (int l, int r) {
    return 1ll * (l + r) * (r - l + 1) / 2;
  };
  int m;
  cin >> m;
  while (m--) {
    int op;
    cin >> op;
    if (op == 1) {
      int u, v;
      cin >> u >> v;
      if (dfn[v] >= dfn[u] && dfn[v] <= dfn[u] + sz[u] - 1) {
        s1 += sz[v];
        s2 += sdp[dfn[v] + sz[v] - 1] - sdp[dfn[v] - 1];
        sgt.modify(dfn[v], dfn[v] + sz[v] - 1, Tag(1, 0, 0));
      } else if (dfn[u] >= dfn[v] && dfn[u] <= dfn[v] + sz[v] - 1) {
        int w = find(u, v);
        s1 += n - sz[w];
        s2 += sdp[n] - sdp[dfn[w] + sz[w] - 1] + sdp[dfn[w] - 1];
        sgt.modify(1, n, Tag(1, 0, 0));
        sgt.modify(dfn[w], dfn[w] + sz[w] - 1, Tag(-1, 0, 0));
        mPath(1, v, Tag(0, 0, -sz[w]));
      } else {
        s1 += sz[v];
        s2 += sdp[dfn[v] + sz[v] - 1] - sdp[dfn[v] - 1];
        sgt.modify(dfn[v], dfn[v] + sz[v] - 1, Tag(1, 0, 0));
      }
    } else if (op == 2) {
      int u, v;
      cin >> u >> v;
      int p = lca(u, v);
      s1 += dep[u] + dep[v] - 2 * dep[p] + 1;
      s2 += calc(dep[p], dep[u]) + calc(dep[p], dep[v]) - dep[p];
      mPath(u, p, Tag(0, -1, dep[u] + 1));
      mPath(v, p, Tag(0, -1, dep[v] + 1));
      mPath(p, p, Tag(0, 0, -1));
      if (p != 1) mPath(1, fa[p], Tag(0, 0, dep[u] + dep[v] - 2 * dep[p] + 1));
    } else {
      int u;
      cin >> u;
      cout << s1 * dep[u] + s2 - 2 * qPath(u) << '\n';
    }
  }
  return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 4ms
memory: 39976kb

input:

5
4 2
2 5
1 5
1 3
5
2 2 4
3 4
2 1 5
2 5 5
3 2

output:

1
5

result:

ok 2 number(s): "1 5"

Test #2:

score: -100
Time Limit Exceeded

input:

200
171 114
50 183
28 68
67 152
139 125
67 55
50 98
106 71
46 42
157 165
42 49
113 12
81 145
105 13
38 96
34 156
24 17
21 191
135 54
174 116
177 157
123 71
95 130
135 193
150 129
25 190
96 93
188 173
90 160
86 187
20 132
199 75
59 195
189 24
40 68
163 83
25 13
73 33
59 50
154 19
146 21
151 67
89 69
...

output:


result: