[省選聯考 2024] 迷宮守衛

二分+貪心+DP。跟 D1T2 思路有點類似，反正很簡單。

複雜度大約是 \({\rm O}(n^22^n)\)。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
const ll inf = 1e18;
int T, n, q[N]; ll K, w[N];
vector<int> Q;
ll dp(int o, int d, int ub) {
    if (d == 0)
        return q[o ^ (1 << n)] <= ub ? inf : 0;
    ll left = dp(o << 1, d - 1, ub), right = dp(o << 1 | 1, d - 1, ub);
    return left + min(w[o], right);
}
void solve(int o, int d, ll &k) {
    if (d == 0) {
        Q.push_back(q[o ^ (1 << n)]);
        return;
    }
    int l = 1, r = 1 << n;
    ll delta = 0;
    while (l < r) {
        int mid = l + r >> 1;
        ll cost = dp(o, d, mid);
        if (cost <= k) l = mid + 1, delta = cost;
        else r = mid;
    }
    Q.push_back(l);
    k -= delta;
    int u = o << d;
    while (q[u ^ (1 << n)] != l) u++;
    for (int i = 0; i < d; i++, u >>= 1) {
        ll cost = dp(u ^ 1, i, l);
        if (u & 1)
            k += cost;
        else {
            k += min(cost, w[u >> 1]);
            if (k < cost) k -= w[u >> 1];
        }
        solve(u ^ 1, i, k);
    }
}
int main() {
    scanf("%d", &T);
    while (T--) {
        scanf("%d%lld", &n, &K);
        for (int i = 1; i < 1 << n; i++)
            scanf("%lld", &w[i]);
        for (int i = 0; i < 1 << n; i++)
            scanf("%d", &q[i]);
        Q.clear();
        solve(1, n, K);
        for (int v : Q) printf("%d ", v);
        putchar('\n');
    }
    return 0;
}