POJ 2479 Maximum sum（动态规划、多段最大子序列求和）

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题目链接：http://poj.org/problem?id=2479

Description

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

$$d(A)=\max_{1\leq s1\leq t1<s2\leq t2\leq n}\{\sum_{i=s1}^{t1}a_i+\sum_{j=s2}^{t2}a_j\}$$
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.

Source

POJ Contest,Author:Mathematica@ZSU

解题思路

题目意思很简单，就是求一段序列的两段最大子序列的和。这是对序列的一段最大子序列求和的推广，并且还能推广到n段子序列求和。核心的思想还是动态规划，定义两个函数：f(i,j)表示以第i个元素结尾的j段子序列最大和，g(i,j)表示前i个元素（不一定以i结尾）的j段子序列最大和。最后的结果就是g(n,2)了。两者的递推关系如下：

$$f(i,j)=max(f(i-1,j)+a_i,g(i-1,j-1)+a_i)$$
$$g(i,j)=max(f(i,j),g(i-1,j))$$
初始化的时候要注意把f和g的各个元素设为负无穷。

AC代码

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
const int num=50005;
int n,t,a[num],f[num][3],g[num][3];

int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n;i++){
            for(int j=1;j<3;j++){
                f[i][j]=g[i][j]=-INF;
            }
        }
        f[1][1]=g[1][1]=a[1];
        for(int i=2;i<=n;i++){
            for(int j=1;j<3;j++){
                f[i][j]=max(f[i-1][j]+a[i],g[i-1][j-1]+a[i]);
                g[i][j]=max(f[i][j],g[i-1][j]);
            }
        }
        printf("%d\n",g[n][2]);
    }
    return 0;
}

后记

这个方法同样适用于多段子序列最大和问题。