ABC260 E - At Least One(双指针)

Posted 2022-08-06 Herio

ABC260 E - At Least One(双指针)

一开始想到two pointers 了，但是不知道怎么快速维护是否满足条件。

原来开一个vector，然后用cnt变量存当前达到的组数。然后只对变为的桶影响cnt。

因为每个

因此复杂度是：

#include <iostream>
#include <vector>
using namespace std;

int main() 
  int N, M;
  cin >> N >> M;
  vector<int> A(N), B(N);
  for (int i = 0; i < N; i++) cin >> A[i] >> B[i];
  vector<vector<int>> inv(M + 1);
  for (int i = 0; i < N; i++) 
    inv[A[i]].push_back(i);
    inv[B[i]].push_back(i);
  
  vector<int> cnt(N), ans(M + 3);
  int cnt_zero = N;
  for (int i = 1, j = 1; i <= M;) 
    while (j <= M and cnt_zero != 0) 
      for (auto& x : inv[j]) 
        if (cnt[x] == 0) cnt_zero--;
        cnt[x]++;
      
      j++;
    
    if (cnt_zero != 0) break;
    ans[j - i]++, ans[M + 1 - i + 1]--;
    for (auto& x : inv[i]) 
      cnt[x]--;
      if (cnt[x] == 0) cnt_zero++;
    
    i++;
  
  for (int i = 1; i <= M; i++) 
    ans[i] += ans[i - 1];
    cout << ans[i] << " \\n"[i == M];
  

---

根据双指针，我们可以枚举左端点，右端点其实我们只需要更新

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