Maximum sum
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 33918 | Accepted: 10504 |
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.
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
立即现场赛了。。3个人尽然没有会dp的sad。。
我仅仅有临阵磨枪了。
题意：给一个数列，求数列中不相交的两个子段和。要求和最大。
线性dp：线性dp的子状态与父状态一般相差一个元素，所以子问题通过加入一个增量而到达父状态。从最小的子问题到原问题。一层一层的状态转移呈现出线性递增的关系。所以称为线性dp。
题解:对于对于每一个状态i。求出[0,i-1]的最大子段和以及[i,n-1]的最大子段和 相加求最大的就可以。[0,i-1]从左往右扫描，[i，n-1]从右往左扫描。
#include <algorithm>
#include <cstdio>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 50010;
#define LL long long
int a[maxn],left[maxn],right[maxn];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
left[0]=a[0];
for(int i=1;i<n;i++)
left[i]=left[i-1]<0?a[i]:left[i-1]+a[i];
for(int i=1;i<n;i++)
left[i]=max(left[i-1],left[i]);
right[n-1]=a[n-1];
for(int i=n-2;i>=0;i--)
right[i]=right[i+1]<0?a[i]:right[i+1]+a[i];
for(int i=n-2;i>=0;i--)
right[i]=max(right[i+1],right[i]);
int ans=-INF;
for(int i=1;i<n;i++)
ans=max(ans,left[i-1]+right[i]);
printf("%d
",ans);
}
return 0;
}