2015 Multi-University Training Contest 1

 Assignment

Problem's Link:  http://acm.hdu.edu.cn/showproblem.php?pid=5289

---

 

Mean: 

给你一个数列和一个k，求连续区间的极值之差小于k的数的个数。

analyse:

用两个优先队列来维护区间的最大值和最小值，每次插入新值的时候检查区间内的极值差是否满足条件，不满足就将最左边的数删除，直到满足条件为止。ans每次加上区间的长度即得最终答案。

Time complexity: O(N)

 

Source code: 

/** this code is made by crazyacking* Verdict: Accepted* Submission Date: 2015-07-21-21.38* Time: 0MS* Memory: 137KB*/#include <queue>#include <cstdio>#include <set>#include <string>#include <stack>#include <cmath>#include <climits>#include <map>#include <cstdlib>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#define  LL long long#define  ULL unsigned long longusing namespace std;int main(){      ios_base::sync_with_stdio( false );      cin.tie( 0 );      int Cas;      cin >> Cas;      while( Cas-- )      {            int n, k, tmp;            cin >> n >> k;            vector<int> v;            for( int i = 0; i < n; ++i )            {                  cin >> tmp;                  v.push_back( tmp );            }            multiset<pair<int, int> > q1, q2;            int l = 0, r = 0;            LL ans = 0;            while( r < n )            {                  q1.insert( make_pair( v[r], r ) ), q2.insert( make_pair( -v[r], r ) );                  while( -( *q1.begin() ).first - ( *q2.begin() ).first >= k )                  {                        q1.erase( make_pair( v[l] , l ) ), q2.erase( make_pair( -v[l], l ) );                        l++;                  }                  ans += r - l + 1;                  r++;            }            cout << ans << endl;      }      return 0;}/**/