POJ2479

Description

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.

题意：求两个不相交串的和的最大值

思路：利用前缀和，从左到右先记录[1,i]序列中的和的最大值，然后从右往左再找和的最大值

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
using namespace std;
const int INF = 1<<29;
const int MAX = 50050;
int n;
int dp[MAX];//dp[i]表示范围在[1,i]内的最大的子段的和
int A[MAX];

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        int tmp = -INF;
        int sum = 0;

        //从左到右找最大子段
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &A[i]);
            sum += A[i];
            if (sum > tmp)
                tmp = sum;
            dp[i] = tmp;
            if (sum < 0)
                sum = 0;
        }

        tmp = -INF; sum = 0;
        int ans = -INF;
        //从右到左找最大子段
        for (int i = n; i >= 2; i--)
        {
            sum += A[i];
            if (sum > tmp)
                tmp = sum;
            if (sum < 0)
                sum = 0;
            if (tmp + dp[i-1] > ans)
                ans = tmp + dp[i-1];
        }
        printf("%d\n", ans);
    }

    return 0;
}