Submission #997543

#TimeUsernameProblemLanguageResultExecution timeMemory
997543biankClosing Time (IOI23_closing)C++17
75 / 100
337 ms13652 KiB
#include <bits/stdc++.h>

using namespace std;

#define sz(x) int(x.size())
#define all(x) begin(x), end(x)

using ll = long long;
using ii = pair<int, int>;

const ll INF = 1e18;
const int MAX_N = 3000;

vector<ii> adj[MAX_N];
pair<ll, ll> val[MAX_N];
int n;

void dfs(int u, vector<ll> &dist, int p = -1, ll d = 0) {
    dist[u] = d;
    for (auto [v, w] : adj[u]) {
        if (v != p) dfs(v, dist, u, d + w);
    }
}

int non_overlaping(ll K) {
    vector<ll> v(n);
    for (int i = 0; i < n; i++) {
        v[i] = val[i].first;
    }
    sort(all(v));
    int ans = 0;
    for (int i = 0; i < n; i++) {
        if (v[i] <= K) {
            K -= v[i];
            ans++;
        }
    }
    return ans;
}

int max_score(int N, int X, int Y, ll K, vector<int> U, vector<int> V, vector<int> W) {
    n = N;
    for (int i = 0; i < n; i++) adj[i].clear(); 
    for (int i = 0; i < n - 1; i++) {
        adj[U[i]].emplace_back(V[i], W[i]);
        adj[V[i]].emplace_back(U[i], W[i]);
    }
    vector<ll> distToX(n), distToY(n);
    dfs(X, distToX), dfs(Y, distToY);
    for (int i = 0; i < n; i++) {
        val[i] = minmax(distToX[i], distToY[i]);
    }
    vector<bool> path(N, false);
    path[Y] = true;
    int u = Y;
    while (u != X) {
        for (auto [v, w] : adj[u]) {
            if (distToX[v] + w == distToX[u]) {
                u = v;
                break;
            }
        }
        path[u] = true;
    }
    vector<ll> prevDP(2 * n + 1, INF);
    prevDP[0] = 0;
    for (int i = 0; i < n; i++) {
        vector<ll> currDP(2 * n + 1, INF);
        if (!path[i]) currDP = prevDP;
        for (int j = 1; j <= 2 * n; j++) {
            currDP[j] = min(currDP[j], prevDP[j - 1] + val[i].first);
        }
        for (int j = 2; j <= 2 * n; j++) {
            currDP[j] = min(currDP[j], prevDP[j - 2] + val[i].second);
        }
        prevDP = currDP;
    }
    for (int i = 2 * n; i >= 0; i--) {
        if (prevDP[i] <= K) return max(i, non_overlaping(K));
    }
    return non_overlaping(K);
}
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