Submission #770546

#	Time	Username	Problem	Language	Result	Execution time	Memory
770546		MohamedFaresNebili	Paths (RMI21_paths)	C++14	48 / 100	1033 ms	35168 KiB

Main

#include <bits/stdc++.h>

        using namespace std;
        using ll = long long;

        #define int ll

        int N, K, P[100005], rem[100005];
        int par[100005][20];
        ll ST[400005], lazy[400005];
        ll D[100005];
        int X[100005], Y[100005], C[100005];
        int timer, tin[100005], out[100005], id[100005];
        ll sol[100005];
        vector<int> adj[100005];

        void update(int v, int l, int r, int p, ll val) {
            if(l == r) {
                ST[v] = val;
                lazy[v] = 0;
                return;
            }
            int md = (l + r) / 2;
            if(p <= md) update(v * 2, l, (l + r) / 2, p, val);
            else update(v * 2 + 1, (l + r) / 2 + 1, r, p, val);
            ST[v] = max(ST[v * 2], ST[v * 2 + 1]);
            lazy[v] = 0;
        }
        void prop(int v, int l, int r) {
            if(l == r || lazy[v] == 0) return;
            lazy[v * 2] += lazy[v];
            lazy[v * 2 + 1] += lazy[v];
            ST[v * 2] -= lazy[v];
            ST[v * 2 + 1] -= lazy[v];
            lazy[v] = 0;
        }
        void update(int v, int l, int r, int lo, int hi, ll val) {
            if(l > hi || r < lo) return;
            if(l >= lo && r <= hi) {
                ST[v] -= val; lazy[v] += val;
                ///prop(v, l, r);
                return;
            }
            prop(v, l, r);
            update(v * 2, l, (l + r) / 2, lo, hi, val);
            update(v * 2 + 1, (l + r) / 2 + 1, r, lo, hi, val);
            ST[v] = max(ST[v * 2], ST[v * 2 + 1]);
        }
        int query(int v, int l, int r) {
            prop(v, l, r);
            if(l == r) return l;
            if(ST[v * 2] >= ST[v * 2 + 1])
                return query(v * 2, l, (l + r) / 2);
            return query(v * 2 + 1, (l + r) / 2 + 1, r);
        }
        void dfs(int v, int p, ll dep = 0) {
            par[v][0] = p; id[timer] = v;
            tin[v] = timer++; update(1, 0, N - 1, tin[v], dep);
            for(auto u : adj[v]) {
                int nxt = ((X[u] ^ Y[u]) ^ v);
                if(nxt == p) continue;
                P[nxt] = u; dfs(nxt, v, dep + C[u]);
            }
            out[v] = timer - 1;
        }
        void solve(int v, int p) {
            sol[v] = ST[1];
            for(auto u : adj[v]) {
                int nxt = ((X[u] ^ Y[u]) ^ v);
                if(nxt == p) continue;
                update(1, 0, N - 1, 0, N - 1, -C[u]);
                update(1, 0, N - 1, tin[nxt], out[nxt], 2 * C[u]);
                solve(nxt, v);
                update(1, 0, N - 1, 0, N - 1, C[u]);
                update(1, 0, N - 1, tin[nxt], out[nxt], -2 * C[u]);
            }
        }

        int32_t main() {
            ios_base::sync_with_stdio(0);
            cin.tie(0); cout.tie(0);
            cin >> N >> K;
            for(int l = 0; l < N - 1; l++) {
                cin >> X[l] >> Y[l] >> C[l];
                int U = X[l], V = Y[l];
                adj[U].push_back(l);
                adj[V].push_back(l);
            }
            if(K == 1) {
                dfs(1, 1);
                solve(1, 1);
                for(int l = 1; l <= N; l++)
                    cout << sol[l] << "\n";
                return 0;
            }
            for(int l = 1; l <= N; l++) {
                for(int i = 1; i <= N; i++)
                    rem[i] = 0;
                rem[l] = 1;
                ll res = 0; timer = 0; dfs(l, l);
                for(int i = 1; i < 11; i++)
                    for(int j = 1; j <= N; j++)
                        par[j][i] = par[par[j][i - 1]][i - 1];
                for(int i = 0; i < K; i++) {
                    int A = query(1, 0, N - 1); res += ST[1]; A = id[A];
                    ///cout << ST[1] << " ";
                    int B = A;
                    for(int j = 10; j >= 0; j--) {
                        if(rem[par[B][j]] == 0)
                            B = par[B][j];
                    }
                    B = par[B][0];

                    while(A != B) {
                        update(1, 0, N - 1, tin[A], out[A], C[P[A]]);
                        rem[A] = 1; A = par[A][0];
                    }
                }
                cout << res << "\n";
            }
        }

• Subtask 1: 8 / 8
• Subtask 2: 11 / 11
• Subtask 3: 17 / 17
• Subtask 4: 0 / 20
• Subtask 5: 12 / 12
• Subtask 6: 0 / 32