poj2593 Max Sequence（求两个不相交最大字段和）

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我是靠谱客的博主无私秋天，最近开发中收集的这篇文章主要介绍poj2593 Max Sequence（求两个不相交最大字段和），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

转载请注明出处：http://blog.csdn.net/u012860063?viewmode=contents

题目链接：http://poj.org/problem?id=2593

Description

Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N).

You should output S.

Input

The input will consist of several test cases. For each test case, one integer N (2 <= N <= 100000) is given in the first line. Second line contains N integers. The input is terminated by a single line with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5
-5 9 -5 11 20
0

Sample Output

思想：对于数据a[]，从左向右依次求解以a[i]结尾的最大子段和b[i]，

然后，从右向左遍历，求a[i]右边（包括a[i]）的最大子段和sum，输出sum+b[i-1]的最大值。

代码如下：

#include <iostream>
using namespace std;
#define INF 0x3fffffff
#define M 100000+17
int a[M],b[M];
int main()
{
	int n,i;
	while(cin >> n && n)
	{
		int sum = 0, MAX = -INF;
		for(i = 1; i <= n; i++)
		{
			cin >> a[i];
			sum+=a[i];
			if(sum > MAX)
			{
				MAX = sum;
			}
			b[i] = MAX;
			if(sum < 0)
			{
				sum = 0;
			}
		}
		MAX = -INF;
		sum = 0;
		int ans = MAX, t;
		for(i = n; i > 1; i--)
		{
			sum+=a[i];
			if(sum > MAX)
			{
				MAX = sum;
			}
			t = MAX + b[i-1];
			if(t > ans)
			{
				ans = t;
			}
			if(sum < 0)
			{
				sum = 0;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}