S - Maximum sum(dp)

叁歲伎倆 2022-02-20 02:58 97阅读 0赞

### 题目描述： ###

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： #####

给定一个序列，求你把序列分成两部份时的和最大值。　  
分析：  
对于每一个i而言，都有\[0, i ) 和\[i, n ) 这两段 。我们分别求出每个 i 对应这两段的最大字段和，然后遍历所有的i，得到最大值即可

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
    using namespace std;
    typedef long long ll;
    ll a[50010];
    ll pre[50010];
    ll lat[50010];
    ll las[50010];
    int main()
    {
        int T;
        int n;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&n);
            for(int i=1; i<=n; i++)
                scanf("%lld",&a[i]);
            memset(pre,0,sizeof(pre));
            memset(lat,0,sizeof(lat));
            memset(las,0,sizeof(las));
            pre[0]=0;
            for(int i=1; i<=n; i++)
                pre[i]=max(a[i],pre[i-1]+a[i]);
            ll maxx=a[n];//同样这里初始值不能为0，有可能值一开始就是小于0的。
            lat[n+1]=0;
            for(int i=n; i>=1; i--)
            {
                lat[i] = max(a[i],lat[i+1]+a[i]);
                maxx = max(maxx,lat[i]);
                las[i] = max(lat[i],maxx);
            }
            maxx=pre[1]+las[2];//初始值不能为0，因为值是有可能小于0的。
            for(int i=1; i<n; i++)
            {
                maxx=max(maxx,pre[i]+las[i+1]);
            }
            printf("%lld\n",maxx);
        }
    }