Monotone beacon transmission queue optimization dp []

Subject to the effect:

1. gives the length n of the array, required to be selected as a value between the cost of every m successive elements, find the minimum cost of the array.

Problem solution ideas:

1. apparently linear dp, dp [i] represents a minimum total consideration when selecting the i-th element. Clearly state transition equation of dp [i] = min (dp [j]) + a [i]. (I - m <= j < = i - 1). However, when evaluated min (dp [j]), we need to iterate over a length of m size range, m and n are as large limits. Time of charge of 0 (N ^ 2). Obviously time out.

2. We need monotone queue length is the minimum to maintain the size of the interval m, the minimum HOL namely, direct O (1) to give min (dp [j]), it can be avoided timeout.

code show as below:

. 1 #include <stdio.h>
 2 #include <the deque>
 . 3 #include <algorithm>
 . 4  const  int MAXN 2E5 + = 100 ;
 . 5  the using  namespace STD;
 . 6  const  int INF = 0x3f3f3f3f ;
 . 7  
. 8  int n-, m, A [ MAXN];
 . 9  int DP [MAXN]; // represents a minimum total cost of the i-th beacon when the beacon is placed 
10 the deque < int > Q;
 . 11  
12 is  int main ()
 13 is  {
 14      Scanf ( " % D% D" , & N-, & m);
 15      for ( int I = . 1 ; I <= n-; I ++ )
 16          Scanf ( " % D " , & A [I]);
 . 17      for ( int I = . 1 ; I <= m; I ++)   // DP initialization and mono queue 
18 is      {
 . 19          DP [I] = A [I];
 20 is          the while (! Q.empty ())
 21 is          {
 22 is              IF (DP [I] <DP [Q. Back ()]) // assurance team is stored in the head dp value of the minimum index of the tail to the head-of monotonically increasing 
23 is                  Q.pop_back ();
24              the else 
25                  BREAK ;
 26 is          }
 27          Q.push_back (I);
 28      }
 29      for ( int I = m + . 1 ; I <= n-; I ++ )
 30      {
 31 is          the while (! Q.empty ())
 32          {
 33 is              IF (I - m> Q.front ()) // guaranteed queue head is within the range of enumeration, i.e. in [i - m, i - 1 ] in the range of 
34 is                  Q.pop_front ();
 35              the else 
36                  BREAK ;
 37 [          }
 38 is          DP [I] DP = [Q.front ()] + a[i];
39         while(!Q.empty())
40         {
41             if(dp[i] < dp[Q.back()])
42                 Q.pop_back();
43             else
44                 break;
45         }
46         Q.push_back(i);
47     }
48     int ans = inf;
49     for(int i = n; i > n - m; i --)
50         ans = min(ans, dp[i]);
51     printf("%d\n", ans);
52     return 0;
53 }
54 /*
55 5 3
56 1 2 5 6 2
57 
58 4
59 */ 

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