Submission #727342

#	Time	Username	Problem	Language	Result	Execution time	Memory
727342		Nelt	Race (IOI11_race)	C++17	100 / 100	506 ms	117684 KiB

race

#include "race.h"

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

/* DEFINES */
#define S second
#define F first
#define ll long long
#define ull unsigned long long
#define ld long double
#define npos ULLONG_MAX
#define INF LLONG_MAX
#define vv(a) vector<a>
#define pp(a, b) pair<a, b>
#define pq(a) priority_queue<a>
#define qq(a) queue<a>
#define ss(a) set<a>
#define mm(a, b) map<a, b>
#define ump(a, b) unordered_map<a, b>
#define sync                      \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define elif else if
#define endl "\n"
#define allc(a) begin(a), end(a)
#define all(a) a, a + sizeof(a) / sizeof(a[0])
#define pb push_back
#define logi(a) __lg(a)
#define sqrt(a) sqrtl(a)
#define mpr make_pair
#define ins insert
using namespace std;
using namespace __gnu_pbds;
using namespace __cxx11;
typedef char chr;
typedef basic_string<chr> str;
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll N = 2e5 + 5;
vv(pp(ll, ll)) g[N];
ll depth[N], depth1[N], ans[N], k;
multiset<pp(ll, ll)> sub[N];
void init(ll v, ll par)
{
    ans[v] = INF;
    for (auto [to, w] : g[v])
        if (to != par)
            depth[to] = depth[v] + 1, depth1[to] = depth1[v] + w, init(to, v);
}
void dfs(ll v, ll par)
{
    sub[v].ins(mpr(depth1[v], depth[v]));
    multiset<pp(ll, ll)>::iterator it;
    for (auto [to, w] : g[v])
    {
        if (to == par)
            continue;
        dfs(to, v);
        if (sub[to].size() > sub[v].size())
            swap(sub[to], sub[v]);
        ans[v] = min(ans[v], ans[to]);
        for (auto [dis, dep] : sub[to])
        {
            it = sub[v].lower_bound(mpr(k - (dis - depth1[v] * 2), -100));
            if (it != sub[v].end() and it->F + dis - depth1[v] * 2 == k)
                ans[v] = min(ans[v], it->S + dep - depth[v] * 2);
        }
        for (auto i : sub[to])
            sub[v].ins(i);
    }
}
int best_path(int n, int k1, int e[][2], int w[])
{
    k = k1;
    for (ll i = 0; i < n - 1; i++)
        g[e[i][0]].pb(mpr(e[i][1], w[i])), g[e[i][1]].pb(mpr(e[i][0], w[i]));
    init(0, 0);
    dfs(0, 0);
    return ans[0] == INF ? -1 : ans[0];
}

• Subtask 1: 9 / 9
• Subtask 2: 12 / 12
• Subtask 3: 22 / 22
• Subtask 4: 57 / 57