【動態規劃】Maximum sum

最新推荐文章于 2024-03-05 20:59:03 发布

Whjpji

最新推荐文章于 2024-03-05 20:59:03 发布

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分类专栏： OI 文章标签： integer input each file output function

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本文介绍了一种解决特定子序列和问题的方法，通过转化问题为求取前缀和后缀的最大子序列和来简化问题。提供了完整的代码实现，并解释了算法思路。

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Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

Your task is to calculate d(A).

Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output
Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

一道簡單的子區間最大和問題。
作一個小小的轉化：題目中要求兩個子序列，且它們的和最大，則轉化為求前綴子序列和後綴子序列最大，這樣求兩次最大子序列就可以將問題解決了。
Accode：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <bitset>

using std::max;
using std::min;

const char fi[] = "poj2479.in";
const char fo[] = "poj2479.out";
const int maxN = 50010;
const int MAX = 0x3fffff00;
const int MIN = -MAX;

int data[maxN];
int f1[maxN], f2[maxN];
int n, N;

  void init_file()
  {
    freopen(fi, "r", stdin);
    freopen(fo, "w", stdout);
  }
  
  void work()
  {
    int tmp = data[1];
    f1[1] = data[1];
    for (int i = 2; i < n + 1; ++i)
    {
      tmp = max(tmp, 0);
      f1[i] = tmp += data[i];
      f1[i] = max(f1[i], f1[i - 1]);
    }
    f2[n] = tmp = data[n];
    for (int i = n - 1; i > 0; --i)
    {
      tmp = max(tmp, 0);
      f2[i] = tmp += data[i];
      f2[i] = max(f2[i], f2[i + 1]);
    }
  }
  
  void readdata()
  {
    scanf("%d", &N);
    for (int t = 0; t < N; ++t)
    {
      memset(f1, 0, sizeof(f1));
      memset(f2, 0, sizeof(f2));
      scanf("%d", &n);
      for (int i = 1; i < n + 1; ++i)
        scanf("%d", data + i);
      work();
      int ans = MIN;
      for (int i = 1; i < n; ++i)
        ans = max(ans, f1[i] + f2[i + 1]);
      printf("%d\n", ans);
    }
  }
  
int main()
{
  init_file();
  readdata();
  exit(0);
}