求数组连续最大和问题

给定一个数组，求解其中连续的最大的子数组

如：1 -1 2 2 3 -3 4 -4 5 -5，最大的连续子数组是：2 2 3 -3 4 -4 5，当然前面的1和－1可要，可不要，得到的结果是9

用程序怎么实现呢，其实这是简单的动态规划题型：

#include<iostream> using namespace std; #define MAX(x,y) x>y?x:y int main() { int arr[] = {-100,50,23,34,-300,34,634 ,3,-1000}; int n = 9; //数组长度 int sum = 0,max = 1<<31; for(int i = 0; i < n; i ++) { sum+=arr[i]; max = MAX(sum,max); if(sum < 0) sum = 0; } cout<<max<<endl; return 0; }  

如果求两个最大的子数组呢？

poj2479就是这样的题目：

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21760		Accepted: 6559

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.

 

这时，我们可以从两头各找出最大，并把每一步的最大记录在数组中，然后在不交叉的条件下求出两个的最大和。如下：

#include <cstdio> #include <iostream> #include <limits.h> using namespace std; int main(int argc, char** argv) { int t,m,i,a[50005],f[2][50005]; scanf("%d",&t); while(t--) { scanf("%d",&m); int sum=0,Max=INT_MIN; for(i=1;i<=m;++i)scanf("%d",&a[i]); for(i=1;i<=m;++i) { sum+=a[i]; Max=Max>sum?Max:sum; if(sum<0)sum=0; f[0][i]=Max; } for(i=m,sum=0,Max=INT_MIN;i>0;--i) { sum+=a[i]; Max=Max>sum?Max:sum; if(sum<0)sum=0; f[1][i]=Max; } for(i=1,sum=INT_MIN;i<m;++i) { sum=max(sum,f[0][i]+f[1][i+1]); } printf("%d/n",sum); } return 0; }