1051. 最大的和

Powered by:NEFU AB-IN

Link

1051. 最大的和

• 题意

  对于给定的整数序列 A={a1,a2,…,an}，找出两个不重合连续子段，使得两子段中所有数字的和最大。

• 思路

  首先是最大子段和的dp证明
  
  最后求出来的dp[i]是以i为结尾的所有连续子序列的和的最大值，而我们要统计整个数组的连续子序列最大值，所以要求${\max }_{i=0}^{n}dp[i]$

  ---

  回到本题
  运用前后缀分解的思想，由于要两段，所以一定存在一个分界点k可以将两段分开
  所以我们需枚举所有的分界点k，求1到k的最大子段和和k+1到n的最大子段和的加和的最大值即可

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2023-03-23 21:42:54
  FilePath: \Acwing\1051\1051.py
  LastEditTime: 2023-03-23 21:54:54
  '''
  read = lambda: map(int, input().split())
  from collections import Counter, deque
  from heapq import heappop, heappush
  from itertools import permutations
  
  N = int(5e4 + 10)
  INF = int(2e9)
  
  dp, g, h, w = [0] * N, [0] * N, [0] * N, [0] * N
  
  for _ in range(int(input())):
      n = int(input())
      w[1:] = list(read())
  
      # s = -INF
      dp[0] = g[0] = -INF  # 这里设成负无穷 是因为每一段都不能是空的，无解就设成-INF
      for i in range(1, n + 1):
          dp[i] = w[i] + max(0, dp[i - 1])
          # s = max(0, s) + w[i]
          g[i] = max(dp[i], g[i - 1])
          # g[i] = max(s, g[i - 1])
  
      # s = -INF
      dp[n + 1] = h[n + 1] = -INF
      for i in range(n, 0, -1):
          dp[i] = w[i] + max(0, dp[i + 1])
          # s = max(0, s) + w[i]
          h[i] = max(dp[i], h[i + 1])
          # h[i] = max(s, h[i + 1])
  
      res = -INF
      for i in range(1, n + 1):
          res = max(res, g[i] + h[i + 1])
      print(res)
  

  ---

  优化版本

  '''
  Author: NEFU AB-IN
  Date: 2023-03-23 21:42:54
  FilePath: \Acwing\1051\1051.py
  LastEditTime: 2023-03-23 21:52:59
  '''
  read = lambda: map(int, input().split())
  from collections import Counter, deque
  from heapq import heappop, heappush
  from itertools import permutations
  
  N = int(5e4 + 10)
  INF = int(2e9)
  
  dp, g, h, w = [0] * N, [0] * N, [0] * N, [0] * N
  
  for _ in range(int(input())):
      n = int(input())
      w[1:] = list(read())
  
      s = -INF
      dp[0] = g[0] = -INF  # 这里设成负无穷 是因为每一段都不能是空的，无解就设成-INF
      for i in range(1, n + 1):
          s = max(0, s) + w[i]
          g[i] = max(s, g[i - 1])
      s = -INF
      dp[n + 1] = h[n + 1] = -INF
      for i in range(n, 0, -1):
          s = max(0, s) + w[i]
          h[i] = max(s, h[i + 1])
  
      res = -INF
      for i in range(1, n + 1):
          res = max(res, g[i] + h[i + 1])
      print(res)