POJ 2479 Maximum sum

题目链接

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32007 Accepted: 9855

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.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1101 -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.

题解：第一次做动态规划题，参照博客的。

1.明白最大连续子段。

2.我感觉做法是的暴力,(1)求1->i所有的最大和，(2)求n->i的最大和，(3)分别优化2个最大和.(4)求2个最大和的最大值即为所求。

AC code：

#include <iostream>#include <cstdio>#include <string>using namespace std;#define max(a,b)  a>b?a:bint main(){int dplift[50001],dpright[50001],i,n,t,sum,a[50001];       scanf("%d",&t);while(t--){//cin>>n;  注意用c语言输入，否则超时scanf("%d",&n);for(i=1;i<=n;i++)//cin>>a[i];scanf("%d",&a[i]);dplift[1]=a[1];for(i=2;i<=n;i++){if(dplift[i-1]>0)dplift[i]=dplift[i-1]+a[i];elsedplift[i]=a[i];}for(i=2; i<=n; i++)dplift[i]=max(dplift[i],dplift[i-1]);dpright[n]=a[n];for(i=n-1;i>0;i--)if(dpright[i+1]>0)dpright[i]=dpright[i+1]+a[i];elsedpright[i]=a[i];for(i = n-1;i>0;i--)dpright[i]=max(dpright[i+1],dpright[i]);sum=dplift[1]+dpright[2];for(i=2; i<n;i++)if(dplift[i]+dpright[i+1]>sum)sum=dplift[i]+dpright[i+1];//cout<<sum<<endl;printf("%d\n",sum);}return 0;}

还有一道类似的注意范围,v_v,被坑了。