poj 2479Maximum sum

最新推荐文章于 2019-01-08 01:46:18 发布

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本文探讨了最大子集和问题的解决方案，通过从左到右及从右到左求解最大连续子数组和的方法，结合动态规划思想，有效地解决了问题。提供了完整的代码实现和样例输入输出。

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Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40740		Accepted: 12706

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

Hint

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

题意：//题目思路是从左到右分别求出它们所在位置的最大连续和，然后从右到左求出它们所在的最大连续和，从后往前求最大和，并加上在此位置之前的一个数的最大和，不断比较

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <set>
using namespace std;
#define N 60000
#define INF 0x3f3f3f3f
int a[N];
int dp[N];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        dp[1]=a[1];
        int sum=a[1];
        int ans=a[1];
        for(int i=2;i<=n;i++) {
            if(sum<0)
                sum=0;
            sum+=a[i];
            if(sum>ans)
                ans=sum;
            dp[i]=ans;
        }
        sum=a[n];
        ans=a[n];
        int maxi=dp[n-1]+sum;
        for(int i=n-1;i>=2;i--) {
            if(sum<0)
                sum=0;
            sum+=a[i];
            if(sum>ans)
                ans=sum;
            maxi=max(maxi,dp[i-1]+ans);
        }
        printf("%d\n",maxi);
    }
    return 0;
}