hdu1024 Max Sum plus Plus

Dynamic programming, given an integer array of length n (≤1e6) and an integer m, selects m consecutive and 22 disjoint sub-ranges, which makes the interval and maximum maximum in all schemes.

DP[I][J] Indicates the end position (the position of the last element in the last interval) is I and the maximum value of the interval number j is selected.

It is easy to get the following state transition equations:

and

Taking into account the size of the array and the updated features of J, a one-dimensional scrolling array is used instead of a two-dimensional array, and the outermost loop enumerates J to M.

Use Dp[0][i] to denote dp[i][j], dp[1][i] for Max (Dp[k][j-1]) (k≤i).

The complexity is O (n^2).

1#include <cstdio>2#include <cstring>3#include <algorithm>4 5 using namespacestd;6 typedef __int64 LL;7 8 Const intINF =0x3f3f3f3f;9 Const intMAXN = 1e6 +Ten;Ten  One intS[MAXN]; ALL dp[2][MAXN]; - LL ans, maxi; - intN, M; the  - intMain () { -      while(~SCANF ("%d%d", &m, &N)) { -          for(inti =1; I <= N; i++) scanf ("%d", &s[i]); +Ans = Maxi =-inf; -Memset (DP,0,sizeofDP); +         into =1; A          for(intj =1; J <= M; J + +){ atMaxi =-inf; -              for(inti = j; I <= N; i++){ -dp[0][i] = S[i] + max (dp[0][i-1], dp[1][i-1]); -             } -              for(inti = j; I <= N; i++) maxi = dp[1][i] = max (maxi, dp[0][i]); -         } in          for(inti = m; I <= N; i++) ans = max (ans, dp[0][i]); -printf"%i64d\n", ans); to     } +     return 0; -}

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hdu1024 Max Sum plus Plus