Bnuoj 27411 heap or monotonic queue maintenance interval RMQ

It's obvious that n-side complexity can't get through.

Consider the optimization of the most value of the query, you can consider using a heap or a monotonous queue to do.

Heap:

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <queue>5 using namespacestd;6 7 Const intINF =99999999;8 Const intN =32768;9 intA[n];Ten intN, l, U; One  A structNode - { -     intl; the     ints; - Node () {} -Node (int_l,int_s) -     { +L =_l; -s =_s; +     } A     BOOL operator< (ConstNode & T)Const at     { -         returns <T.s; -     } - }; -  - intSolve () in { -Priority_queue<node>Q; to     intAns =INF; +      for(inti = l; I <= N; i++ ) -     { theQ.push (Node (i-l, A[i-l])); *          while(Q.top (). L + U <i) Q.pop (); $         intTMP = A[i]-Q.top (). s;Panax Notoginseng         if(tmp < ans) ans =tmp; -     } the     returnans; + } A  the intMain () + { -      while(SCANF ("%d", &N), N) $     { $scanf"%d%d", &l, &u); -a[0] =0; -          for(inti =1; I <= N; i++ ) the         { -scanf"%d", A +i);WuyiA[i] + = a[i-1]; the         } -printf"%d\n", Solve ()); Wu     } -     return 0; About}

Monotonic queue:

Bnuoj 27411 heap or monotonic queue maintenance interval RMQ