Poj 2479-maximum sum (linear DP)

Maximum Sum

Time limit:1000 ms		Memory limit:65536 K
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 ().

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 shoshould contain the integer d ().

Sample Input

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

Sample output

13

马上现场赛了。。3个人尽然没有会dp的sad。。我只有临阵磨枪了。

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，所以子问题通过添加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 相加求最大的即可。[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 longint 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\n",ans);}return 0;}

Poj 2479-maximum sum (linear DP)