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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#338342#7108. CouleurLainML 1ms3544kbC++236.3kb2024-02-25 20:44:552024-02-25 20:44:56

Judging History

你现在查看的是最新测评结果

  • [2024-02-25 20:44:56]
  • 评测
  • 测评结果:ML
  • 用时:1ms
  • 内存:3544kb
  • [2024-02-25 20:44:55]
  • 提交

answer

#include "bits/stdc++.h"
using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;


// Source: Me
// Tested on: ???
struct PersistentSegTree {
  struct SegTreeNode {
    SegTreeNode *l, *r;  // Children

    // Node values
    int sum = 0;

    SegTreeNode(int val = 0) : l(nullptr), r(nullptr), sum(val) {}
    SegTreeNode(SegTreeNode* l, SegTreeNode* r) : l(l), r(r), sum(0) {
      if (l) sum += l->sum;
      if (r) sum += r->sum;
    }
  };

  vector<SegTreeNode*> roots;  // History
  int L, R;                    // Bounds

  PersistentSegTree(int tl, int tr) : L(tl), R(tr) {
    roots.push_back(build(tl, tr));
  }

  int size() { return (int)roots.size(); }

  // Get node from history
  SegTreeNode*& operator[](int idx) {
    assert(idx < roots.size());
    return roots[idx];
  }

  // Build segtree
  SegTreeNode* build(int tl, int tr) {
    if (tl == tr) return new SegTreeNode(0);
    int tm = (tl + tr) / 2;
    return new SegTreeNode(build(tl, tm), build(tm + 1, tr));
  }

  // Query current node for range [l,r]
  int query(int l, int r) { return query(roots.back(), l, r, L, R); }

  // Query node ind for range [l,r]
  int query(int ind, int l, int r) { return query(roots[ind], l, r, L, R); }

  int query(SegTreeNode* v, int l, int r, int tl, int tr) {
    if (l > r) return 0;
    if (l == tl && r == tr) return v->sum;
    int tm = (tl + tr) / 2;
    return query(v->l, l, min(r, tm), tl, tm) +
               query(v->r, max(l, tm + 1), r, tm + 1, tr);
  }

  // Point update position pos to value val
  void update(int pos, int val) {
    roots.push_back(update(roots.back(), pos, val, L, R));
  }

  SegTreeNode* update(SegTreeNode* v, int pos, int val, int tl, int tr) {
    if (tl == tr) return new SegTreeNode(v->sum + val);
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      return new SegTreeNode(update(v->l, pos, val, tl, tm), v->r);
    else
      return new SegTreeNode(v->l, update(v->r, pos, val, tm + 1, tr));
  }
};


// Source: Lain
// Tested on: https://judge.yosupo.jp/problem/point_add_range_sum
//
// Implementation of a Fenwick Tree.  can be used to make
// a order-statistics tree.
template <typename T>
struct Fenwick {
 public:
  Fenwick() = default;
  Fenwick(int n) : n(n), tree(n + 1, 0) {}
  Fenwick(const vector<T>& build) : Fenwick(build.size()) {
    for (int i = 1; i <= n; i++) {
      tree[i] = build[i - 1];
      for (int k = (i & -i) >> 1; k > 0; k >>= 1) tree[i] += tree[i - k];
    }
  }

  void add(int pos, const T& change) {
    assert(pos < n);
    for (int i = pos + 1; i <= n; i += (i & -i)) tree[i] += change;
  }

  T query(int r) {
    assert(r < n);
    T ret = 0;
    for (int i = r + 1; i > 0; i -= (i & -i)) ret += tree[i];
    return ret;
  }

  T query(int l, int r) {
    return (l == 0) ? query(r) : query(r) - query(l - 1);
  }

  // Returns the largest p in [0,tn] such that query(p) <= sum
  int find_last_prefix(T sum) {
    if (sum < 0) return -1;
    int prefix = 0;
    for (int k = 31 - __builtin_clz(n); k >= 0; k--) {
      if (prefix + (1 << k) <= n && tree[prefix + (1 << k)] <= sum) {
        prefix += 1 << k;
        sum -= tree[prefix];
      }
    }
    return prefix;
  }

 private:
  size_t n;
  vector<T> tree;
};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int tt;
  cin >> tt;
  while(tt--) {
    int n;
    cin >> n;
    vi a(n);
    for (auto&x : a) cin >> x, x--;
    vector<int64_t> p(n);
    for (auto& x : p) cin >> x;

    int64_t ans = 0;
    Fenwick<int> T(n);
    PersistentSegTree PT(0, n-1);
    for (int i = 0; i < n; i++) {
      if (a[i]+1 <= n-1) {
        ans += T.query(a[i]+1, n-1);
      }
      T.add(a[i], 1);
      PT.update(a[i], 1);
    }
    for (int i =0; i < n; i++) {
      T.add(a[i], -1);
    }

    set<array<int64_t,3>> segments;
    multiset<int64_t> all_values;
    all_values.insert(ans);
    segments.insert({0, n-1, ans});
    rep(i, 0, sz(p)) {
      int64_t ans = *all_values.rbegin();
      cout << ans << " \n"[i+1 == sz(p)];

      int rem = p[i]^ans;
      rem--;
      auto it = segments.upper_bound({rem, n+10});
      it--; // segment now contains rem
      auto [l, r, val] = *it;
      segments.erase(it);
      all_values.erase(all_values.find(val));

      // remove influence of rem
      if (rem != r && a[rem] - 1 >= 0) {
        // how many values are < a[rem] in range (rem+1, r)?
        val -= PT.query(r+1, 0, a[rem] - 1) - PT.query(rem+1, 0, a[rem] - 1);
      }
      if (rem != l && a[rem]+1 <= n-1) {
        // how many values are > a[rem] in range (l, rem-1)
        val -= PT.query(rem, a[rem] + 1, n-1) - PT.query(l, a[rem] + 1, n-1);
      }

      if (rem-l <= r-rem) {
        // iterate on the left side
        int64_t left_sum = 0;
        rep(j, l, rem) {
          if (a[j] -1 >= 0) {
            val -= PT.query(r+1, 0, a[j] - 1) - PT.query(rem+1, 0, a[j] - 1);
          }
          T.add(a[j], 1);
          if (a[j] + 1 <= n-1) {
            left_sum += T.query(a[j]+1, n-1);
          }
        }
        rep(j, l, rem) {
          T.add(a[j], -1);
        }
        if (l < rem) {
          segments.insert({l, rem-1, left_sum});
          all_values.insert(left_sum);
        }
        if (r > rem) {
          segments.insert({rem+1, r, val-left_sum});
          all_values.insert(val - left_sum);
        }
      } else {
        // iterate on the right side
        int64_t right_sum = 0;
        rep(j, rem+1, r+1) {
          if (a[j] + 1 <= n-1) {
            val -= PT.query(rem, a[j] + 1, n-1) - PT.query(l, a[j] + 1, n-1);
          }
          T.add(a[j], 1);
          if (a[j] + 1 <= n-1) {
            right_sum += T.query(a[j]+1, n-1);
          }
        }
        rep(j, rem+1, r+1) {
          T.add(a[j], -1);
        }
        if (l < rem) {
          segments.insert({l, rem-1, val - right_sum});
          all_values.insert(val - right_sum);
        }
        if (r > rem) {
          segments.insert({rem+1, r, right_sum});
          all_values.insert(right_sum);
        }
      }
    }
  }
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3544kb

input:

3
5
4 3 1 1 1
5 4 5 3 1
10
9 7 1 4 7 8 5 7 4 8
21 8 15 5 9 2 4 5 10 6
15
4 8 8 1 12 1 10 14 7 14 2 9 13 10 3
37 19 23 15 7 2 10 15 2 13 4 5 8 7 10

output:

7 0 0 0 0
20 11 7 2 0 0 0 0 0 0
42 31 21 14 14 4 1 1 1 0 0 0 0 0 0

result:

ok 3 lines

Test #2:

score: -100
Memory Limit Exceeded

input:

11116
10
10 5 10 3 6 4 8 5 9 8
31 27 24 11 12 3 0 2 3 1
10
8 2 7 2 8 10 1 10 9 10
6 5 2 13 2 1 0 1 3 1
10
7 10 7 6 1 3 10 6 7 9
21 18 10 1 6 5 4 8 9 10
10
2 10 4 8 8 5 7 2 6 7
20 10 9 1 15 0 4 2 9 7
10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
10
1 2 3 4 5 6 7 8 9 10
6 3 5 2 7 10 9 1 4 8
10
1 10 1 3...

output:

21 18 16 12 10 6 4 1 1 0
12 12 10 10 4 4 4 2 1 0
20 16 9 5 3 3 3 0 0 0
22 14 8 8 5 5 2 1 1 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
19 12 7 4 4 2 2 1 0 0
20 18 8 3 1 1 0 0 0 0
45 21 21 10 3 3 3 0 0 0
17 11 8 2 1 1 1 0 0 0
13 4 1 0 0 0 0 0 0 0
29 27 22 15 9 7 4 3 1 0
26 16 9 2 1 1 1 1 1 0
0 0 0 0 0 ...

result: