2015 Multi-University Training Contest 1 - 1002 Assignment

 Assignment#

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

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/* * 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 long using 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; } /*   */