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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
#327618#3307. Query on a Tree 17JWRuixiWA 1ms6052kbC++203.7kb2024-02-15 11:30:282024-02-15 11:30:28

Judging History

你现在查看的是最新测评结果

  • [2024-02-15 11:30:28]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:6052kb
  • [2024-02-15 11:30:28]
  • 提交

answer

#include <bits/stdc++.h>
#define IL inline
#define LL long long
#define eb emplace_back
#define L(i, j, k) for (int i = (j); i <= (k); ++i)
#define R(i, j, k) for (int i = (j); i >= (k); --i)
using namespace std;

namespace IO {
  char ibuf[1 << 20], *iS, *iT;
#ifndef LOCAL
#define gh() (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, 1 << 20, stdin), (iS == iT ? EOF : *iS++) : *iS++)
#else
#define gh() getchar()
#endif
  template<typename T = int>
  IL T read () {
    T x = 0;
    bool f = 0;
    char c = gh();
    while (c < '0' || '9' < c) f |= c == '-', c = gh();
    while ('0' <= c && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = gh();
    return f ? ~(x - 1) : x;
  }
}

using IO::read;
using ci = const int;
using vi = vector<int>;

constexpr int N = 1e5 + 9;
constexpr int M = 20;
int n, m, dep[N], sz[N], son[N], top[N], dfn[N], id[N], dfc, par[N], st[N][M];
vi g[N];

namespace SGT {
  LL tr[N << 2], tg[N << 2];

#define ls (p << 1)
#define rs (p << 1 | 1)

  IL void pushup (int p) {
    tr[p] = tr[ls] + tr[rs];
  }

  IL void push (int p, int v) {
    tr[p] += v;
    tg[p] += v;
  }

  IL void pushdown (int p) {
    auto &o = tg[p];
    if (o) {
      push(ls, o);
      push(rs, o);
      o = 0;
    }
  }

  IL void upd (int ql, int qr, int l, int r, int p) {
    if (ql <= l && r <= qr) {
      push(p, 1);
      return;
    }
    pushdown(p);
    ci mid = (l + r) >> 1;
    if (ql <= mid) upd(ql, qr, l, mid, ls);
    if (mid < qr) upd(ql, qr, mid + 1, r, rs);
    pushup(p);
  }

  IL void upd (int l, int r) {
    upd(l, r, 1, n, 1);
  }

  IL LL qry (int ql, int qr, int l, int r, int p) {
    if (ql <= l && r <= qr) {
      return tr[p];
    }
    pushdown(p);
    ci mid = (l + r) >> 1;
    LL R = 0;
    if (ql <= mid) R += qry(ql, qr, l, mid, ls);
    if (mid < qr) R += qry(ql, qr, mid + 1, r, rs);
    return R;
  }

  IL LL qry (int l, int r) {
    return qry(l, r, 1, n, 1);
  }
}

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

IL void dfs2 (int u, int tp) {
  top[u] = tp;
  id[dfn[u] = ++dfc] = u;
  if (son[u]) {
    dfs2(son[u], tp);
  }
  for (int v : g[u]) {
    if (v == son[u] || v == par[u]) continue;
    dfs2(v, v);
  }
}

void init () {
  dfs1(1, 0);
  dfs2(1, 1);
  L (i, 1, n) {
    int k = id[i];
    st[k][0] = par[k];
    L (j, 1, 17) st[k][j] = st[st[k][j - 1]][j - 1];
  }
}

IL LL siz (int u) {
  return SGT::qry(dfn[u], dfn[u] + sz[u] - 1);
}

int main () {
  n = read();
  L (i, 1, n - 1) {
    int u = read(), v = read();
    g[u].eb(v);
    g[v].eb(u);
  }
  init();
  m = read();
  LL all = 0;
  while (m--) {
    int o = read();
    if (o == 1) {
      int x = read();
      SGT::upd(dfn[x], dfn[x] + sz[x] - 1);
      all += sz[x];
    } else {
      int x = read(), y = read();
      all += dep[x] + dep[y] + 1;
      while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        SGT::upd(dfn[top[x]], dfn[x]);
        x = par[top[x]];
      }
      if (dep[x] < dep[y]) swap(x, y);
      SGT::upd(dep[y], dfn[x]);
      all -= 2 * dep[y];
    }
    int l = 1, r = n, md, p = 0;
    while (l <= r) {
      md = (l + r) >> 1;
      SGT::qry(1, md) * 2 >= all ? p = md, r = md - 1 : l = md + 1;
    }
    p = id[p];
    if (siz(p) * 2 > all) {
      printf("%d\n", p);
      continue;
    }
    R (i, 17, 0) {
      int q = st[p][i];
      if (q && siz(q) * 2 <= all) {
        p = st[p][i];
      }
    }
    printf("%d\n", par[p]);
  }
}
// I love WHQ!

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 5984kb

input:

7
1 6
1 7
7 3
3 2
7 5
5 4
4
1 2
1 4
1 6
2 6 7

output:

2
7
7
1

result:

ok 4 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 6052kb

input:

15
1 2
1 3
2 4
3 7
6 2
3 8
4 9
2 5
5 11
7 13
10 4
11 15
12 5
14 7
11
2 6 11
1 6
2 10 9
1 9
2 2 6
1 4
2 7 6
1 2
2 8 13
1 10
2 8 15

output:

2
2
2
2
2
2
2
2
2
2
2

result:

ok 11 lines

Test #3:

score: -100
Wrong Answer
time: 1ms
memory: 5812kb

input:

2
1 2
1
1 1

output:

0

result:

wrong answer 1st lines differ - expected: '1', found: '0'