poj 2479 Maximum sum(前缀和 + dp)

目录

 

题目：

题目大意：

思路：

AC代码：

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题目：

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 43867		Accepted: 13643

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

1

10
1 -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.

题目大意：

给出一串数字，要求从中选出两个部分使得，这两个子串和最大。

思路：

这是一道很经典的dp问题，先顺序遍历一遍，将当前位置的最大前缀和用dp数组记录，然后逆序遍历一遍将当前位置最大前缀和与dp数组加和求最大值

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N  = 1e5 + 7;
int dp[N],a[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n,temp = -inf,maxn = -inf,ans = -inf;
        scanf("%d",&n);
        memset(dp,0,sizeof dp);
        for(int i = 0;i < n;++i) scanf("%d",&a[i]);
        for(int i = 0;i < n;++i){
            temp = temp >= 0 ? temp + a[i] : a[i] ;
            if(temp > maxn){
                maxn = temp;
                dp[i] = maxn;
            }
        }
        temp = maxn = -inf;
        for(int i = n - 1;i > 0;--i){
            temp = temp >= 0 ? temp + a[i] : a[i];
            if(temp > maxn) maxn = temp;
            if(maxn + dp[i - 1] > ans) ans = maxn + dp[i - 1];
        }
        printf("%d\n",ans);
    }
    return 0;
}