POJ 2479

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本文详细阐述了如何解决无交叉的两端序列最大字段和问题，包括输入输出规范、核心思路和代码实现。通过实例分析，帮助读者理解并掌握该算法的应用。

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A - Maximum sum

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2479

Description

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

$d(A)=\max_{1\leq s_1\leq t_1<s_2\leq t_2\leq n}\left\{\sum_{i=s_1}^{t_1}a_i+\sum_{j=s_2}^{t_2}a_j\right\}$

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

Hint

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

Huge input,scanf is recommended.

求一个序列的无交叉的两端序列最大字段和问题。

思路：先从左向右求出各段最大和，再从右向左，求出各段最大和，然后两段无交叉相加求出最大值即可

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 55555
#define MIN -9999999
using namespace std;

int T;
int n;
int a[N];
int num[N];
int ans;

int main()
{

    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%d",&n);
            int sum=0;
            int f=MIN;
            memset(num,0,sizeof num);

            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
                sum+=a[i];

                if(sum>f)
                f=sum;

                num[i]=f;
                if(sum<0)
                sum=0;
            }

            sum=0;
            f=MIN;
            ans=MIN;

            for(int i=n;i>1;i--)
            {
                sum+=a[i];
                if(sum>f)
                f=sum;

                if(ans<num[i-1]+f)
                ans=num[i-1]+f;

                if(sum<0)
                sum=0;
            }
            printf("%d\n",ans);
        }
    }

    return 0;
}