1481:Maximum sum

1481:Maximum sum

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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

                     t1     t2 
         d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }
                    i=s1   j=s2

Your task is to calculate d(A).

输入

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.

输出

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

样例输入

1

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

样例输出

13



//最大连续和的变种，dp数组表示以i结尾的子串最大和，l数组表示i往左最大的一个和，后二者表示反过来

#include<iostream>
#include<memory.h>
using namespace std;
int dp[50001], rdp[50001], l[50001], r[50001], data[50001];
int main() {
    int T; scanf("%d", &T);
    while (T --) {
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; i ++) scanf("%d", &data[i]);
        dp[0] = data[0]; rdp[n - 1] = data[n - 1];
        memset(l, 0, sizeof(l));
        memset(r, 0, sizeof(r));
        l[0] = dp[0]; r[n - 1] = rdp[n - 1];
        for (int i = 1; i < n; i ++) {
            dp[i] = data[i];
            if (dp[i - 1] > 0) dp[i] += dp[i - 1];
            l[i] = max(l[i - 1], dp[i]);
        }
        for (int i = n - 2; i >= 0; i --) {
            rdp[i] = data[i];
            if (rdp[i + 1] > 0) rdp[i] += rdp[i + 1];
            r[i] = max(r[i + 1], rdp[i]);
        }
        int ans = -2147483646;//注意全为负数的情况
        for (int i = 0; i < n - 1; i ++)
            ans = max(ans, l[i] + r[i + 1]);
        printf("%d\n", ans);
    }
}