Yuki 626 Randomized 01 Knapsack

Problem Description:
0/1 Knapsack problem. Test cases are generated at complete randomness.

Constraints:
2 ≤ N ≤ 5000
v[i], w[i] is uniformly distributed in range [1, 1e12]
1 ≤ W ≤ 1e12 * N

Solution:
Branch and Bound. Referenced kimiyuki's blog post.

Code:

#include <bits/stdc++.h>

long long branch_and_bound(int n, long long c, const std::vector<long long> &v, const std::vector<long long> &w) {
  std::vector<long long> vv(n), ww(n); {
    std::vector<int> id(n);
    std::iota(id.begin(), id.end(), 0);
    std::sort(id.begin(), id.end(), [&](int i, int j) {
      return 1.0 * v[i] * w[j] > 1.0 * v[j] * w[i];
    });
    for (int i = 0; i < n; ++i) {
      vv[i] = v[id[i]];
      ww[i] = w[id[i]];
    }
  }
  long long ans = 0;
  std::function<void(int, long long, long long)> dfs = [&](int x, long long sv, long long sw) {
    if (c < sw) return;
    if (x == n) return ans = std::max(ans, sv), void();
    long long gsv = sv, gsw = sw;
    int y = x;
    for ( ; y < n && gsw + ww[y] <= c; ++y) {
      gsv += vv[y];
      gsw += ww[y];
    }
    if (gsw == c || y == n) return ans = std::max(ans, gsv), void();
    if (gsv + 1.0 * vv[y] * (c - gsw) / ww[y] <= ans) return;
    dfs(x + 1, sv + vv[x], sw + ww[x]);
    dfs(x + 1, sv, sw);
  };
  dfs(0, 0, 0);
  return ans;
}

signed main() {
  std::ios::sync_with_stdio(false);
  int N;
  long long C;
  std::cin >> N >> C;
  std::vector<long long> V(N), W(N);
  for (int i = 0; i < N; ++i) {
    std::cin >> V[i] >> W[i];
  }
  std::cout << branch_and_bound(N, C, V, W) << std::endl;
  return 0;
}