poj 2479 max sum

標籤：演算法   poj   

Maximum sum Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmitStatus

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題意分析：求從左至右的最大的連續子序列的和然後在從右至左的一段連續子序列的和，但是這兩段區間不能重疊；簡單的DP：
#include<iostream>#define max(a,b) a>b?a:busing namespace std;int sum[50010],dp1[50010],dp2[50010];int a[50010];int main(){  int i,j,t,n;  int maxn;  int N;  cin>>N;  while(N--)  {   cin>>n;   maxn=-0x3fffffff;   for(i=1;i<=n;i++)   {  scanf("%d",a+i);   }   sum[1]=a[1];   dp1[1]=a[1];   for(i=2;i<=n;i++)   {     sum[i]=max(sum[i-1]+a[i],a[i]); maxn=max(sum[i],maxn); dp1[i]=maxn;   }         maxn=-0x3fffffff;dp2[n]=sum[n]=a[n]; for(i=n-1;i>=1;i--) {   sum[i]=max(sum[i+1]+a[i],a[i]);   maxn=max(maxn,sum[i]);       dp2[i]=maxn; }    maxn=-0x3fffffff;for(i=1;i<n;i++){  int temp=dp1[i]+dp2[i+1];  maxn=max(maxn,temp);}if(N!=1) cout<<endl;cout<<maxn<<endl;  }  return 0;}

poj 2479 max sum