poj-2479 Maximum sum 【最大字串和】

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Maximum sum

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.

Source

POJ Contest ,Author:Mathematica@ZSU

求两段的最大和，前后进行两次dp；

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

int t, n, p[100010], ans[100010];

int main()
{
	scanf("%d",&t);
	while (t--)
	{
		memset(ans,0,sizeof(ans));
		scanf("%d",&n);
		int b = 0, sum = -1000000000;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &p[i]);
			b += p[i];
			sum = max(b, sum);
			ans[i] = sum;
			if (b < 0) 
				b = 0;	
		}

		b = 0, sum = -1000000000;
		int tmp = -1000000000;

		for (int i = n - 1; i >= 1; i--)
		{
			b += p[i];
			sum = max(b, sum);
			tmp = max(tmp, ans[i - 1] + sum);
			if (b < 0) 
				b = 0;
		}
		printf("%d\n",tmp);
	}
	return 0;
}