28:Maximum sum

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

1101 -1 2 2 3 -3 4 -4 5 -5

样例输出

13

提示

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：

就是找到两个字段和，找出最大的和，输出，中间不能有重叠

分析：

就是从左到右找每个i处的最大子段和，在从右到左，最后统计和的最大值

代码：

#include<bits/stdc++.h>using namespace std;long long int f[50005],h[50005],maxn,temp,p;int main(){    int a[50005],n,i,t,k;    while(cin>>t)    {        for(k=0;k<t;k++){        cin>>n;        for(i=1;i<=n;i++)            cin>>a[i];        memset(f,0,sizeof(f));        memset(h,0,sizeof(h));        for(i=1,temp=0,h[0]=-10005;i<=n;i++)        {            if(temp<0)                temp=0;            temp+=a[i];            if(temp>h[i-1])                h[i]=temp;            else            h[i]=h[i-1];        }        for(i=n,temp=0,f[n+1]=-10005;i>0;i--)        {            if(temp<0)                temp=0;            temp+=a[i];            if(temp>f[i+1])                f[i]=temp;            else            f[i]=f[i+1];        }        for(i=1,maxn=-100000;i<n;i++)        {            p=f[i+1]+h[i];            if(p>maxn)                maxn=p;        }        cout<<maxn<<endl;        }    }}

感受：

一开始老是出不来，因为我没有处理好重叠部分的问题，最后想到了算f[i]+h[i-1],不过想出来就感觉自己棒棒哒