QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#82617#2550. Lion and Zebrawoxiangbaile#WA 4ms9496kbC++141.9kb2023-02-28 14:25:482023-02-28 14:25:50

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-02-28 14:25:50]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:9496kb
  • [2023-02-28 14:25:48]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define mp make_pair
#define pb emplace_back
#define rep(i, s, e) for (int i = s; i <= e; ++i)
#define drep(i, s, e) for (int i = s; i >= e; --i)
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define pv(a) cout << #a << " = " << a << endl
#define pa(a, l, r) cout << #a " : "; rep(_, l, r) cout << a[_] << ' '; cout << endl

using pii = pair <int, int>;

const int N = 1e5 + 10;

int read() {
  int x = 0, f = 1; char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  return x * f;
}

int n, q, p[N], ans[N], up[N], down[N], m, a[N];
vector <int> e[N];
vector <pii> qry[N];

void dfs0(int u, int fa) {
  p[u] = fa;
  for (int v : e[u]) if (v != fa) {
    dfs0(v, u), down[u] = max(down[u], down[v] + 1);
  }
}

void dfs1(int u, int fa) {
  int x = 0, y = 0;
  auto upd = [&](int d) {
    if (d > x) y = x, x = d;
    else if (d > y) y = d;
  } ;
  for (int v : e[u]) if (v != fa) upd(down[v] + 1);
  for (int v : e[u]) if (v != fa) {
    up[v] = up[u] + 1;
    if (down[v] + 1 == x) up[v] = max(up[v], y + 1);
    else up[v] = max(up[v], x + 1);
    dfs1(v, u);
  }
}

int main() {
  n = read(), q = read();
  rep(i, 1, n - 1) {
    int u = read(), v = read();
    e[u].pb(v), e[v].pb(u);
  }
  rep(i, 1, q) {
    int u = read(), d = read();
    qry[u].pb(mp(d, i));
  }
  dfs0(1, 0), dfs1(1, 0);
  rep(u, 1, n) if (qry[u].size()) {
    m = 0;
    for (int v : e[u]) a[++ m] = (p[v] == u ? down[v] : up[v]) + 1;
    sort(a + 1, a + m + 1);
    for (auto [d, id] : qry[u]) {
      int deg = m - (lower_bound(a + 1, a + m + 1, d) - a - 1);
      if (d <= 2 * (deg - 1)) ans[id] = d;
      else ans[id] = d + a[m - 1];
    }
  }
  rep(i, 1, q) printf("%d\n", ans[i]);
  return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 4ms
memory: 9496kb

input:

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

output:

4
1

result:

ok 2 number(s): "4 1"

Test #2:

score: 0
Accepted
time: 3ms
memory: 8192kb

input:

11 2
1 2
2 3
1 4
4 5
1 6
6 7
7 8
1 9
9 10
10 11
3 2
10 4

output:

2
5

result:

ok 2 number(s): "2 5"

Test #3:

score: -100
Wrong Answer
time: 2ms
memory: 8252kb

input:

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

output:

4
3

result:

wrong answer 1st numbers differ - expected: '5', found: '4'