Maximum sum

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Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34590 Accepted: 10712

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.

Source

POJ Contest,Author:Mathematica@ZSU

刚开始对于这个题一点思路都木有，还好有大神教，明白了这种题该怎么做。由于这个题是求两个集合的最大值，所以我们可以分别从前边和后边去求每个集合的最大值，然后再去枚举i，看看从哪里断开能使前边的集合最大值加上后边的集合最大值是最大。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <algorithm>#define LL long longusing namespace std;LL dp[2][50400];             //一般动态规划的题都是开longlong整型的LL st[50400];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int i,n;        LL MAX;        scanf("%d",&n);        for(i=1; i<=n; i++)            scanf("%lld",&st[i]);        dp[0][1]=st[1];        for(i=2; i<=n; i++)            dp[0][i]=max(dp[0][i-1],(LL)0)+st[i];         //dp[0][i]代表从前边开始遍历，遍历第一遍，求出1~i的值        for(i=2; i<=n; i++)            dp[0][i]=max(dp[0][i],dp[0][i-1]);            //遍历第二遍，求出1~i的最大值        dp[1][n]=st[n];                                   //dp[1][i]代表从后边开始遍历        for(i=n-1; i>=1; i--)            dp[1][i]=max(dp[1][i+1],(LL)0)+st[i];        for(i=n-1; i>=1; i--)            dp[1][i]=max(dp[1][i],dp[1][i+1]);        MAX=dp[0][1]+dp[1][2];                              //假设从i=1断开是最大        for(i=2; i<n; i++)                                 //开始枚举            MAX=max(MAX,dp[0][i]+dp[1][i+1]);        printf("%lld\n",MAX);    }    return 0;}