제출 #1031453

#제출 시각아이디문제언어결과실행 시간메모리
1031453aaaaaarroz봉쇄 시간 (IOI23_closing)C++17
83 / 100
1051 ms36724 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 = 2e5 + 9;
     
    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);
        bool flag = false;
        for (int i = 0; i < n; i++) {
            val[i] = minmax(distToX[i], distToY[i]);
            if (val[i].second <= K) flag = true;
        }
        if (!flag) return non_overlaping(K);
        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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