动态规划练习28:Maximum sum

题目简要：

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.

这道题就是求最大二子段的问题。

解题思路：

对于每一个数都求一下有这个数的前面的最大子段和，还有后面的最大字段和，最后加起来，放到数组里，求最大就好啦。

附代码：

#include<bits/stdc++.h>
using namespace std;
int a[50005],b[50005],c[50005];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
int sum=0;
b[0]=-100000;
c[n+1]=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i-1]>sum+a[i])
b[i]=b[i-1];
else b[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
sum=0;
for(int i=n;i!=0;i--)
{
if(c[i+1]>sum+a[i])
c[i]=c[i+1];
else c[i]=sum+a[i];
sum=sum+a[i];
if(sum<0)
sum=0;
}
int answer=-1000000;
for(int i=1;i<=n;i++)
{
if(b[i]+c[i+1]>answer)
answer=b[i]+c[i+1];
}
cout<<answer<<endl;
}
return 0;
}

解题感受：

其实这个问题只要想到就好求啦，虽然WR了4次····