最大子数组

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 120366    Accepted Submission(s): 27835

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4 Case 2: 7 1 6

这个是我根据算法导论上写的，大牛们如果看到了请指点一下这个下标该怎么求！！！

 --我已经自己改正了，仅仅改成结构体就可以了！

//最大子数组之和正确，但是下标不正确！！！
/*
测试用例：
9
13 1 2 3 -100 1 2 3 -100 1 2 1 1 1
5 -3 -2 -1 -2 -3 
正确结果：
Case 1： 6 9 13
—— -1 3 3
*/
#include<iostream>
#define MIN -1000000005
using namespace std;

int left_low,left_high,right_low,right_high,cross_low,cross_high,flg;

int max(int a, int b){return a>b?a:b;}

//求过中点的最大和
int maxCenterSum(int *A, int low, int mid, int high)
{
	int left_sum=MIN,sum=0,right_sum=MIN;
	int i,j;

	for(i=mid;i>=low;i--)
	{
		sum=sum+A[i];
		if(sum>=left_sum){left_sum=sum;cross_low=i;}
	}
	sum=0;
	for(j=mid+1;j<=high;j++)
	{
		sum=sum+A[j];
		if(sum>=right_sum){right_sum=sum;cross_high=j;}
	}
	return left_sum+right_sum;
}

//分别左右最大
int maxSub(int *A, int low, int high)
{
	if(low==high)return A[low];
	int mid=(low+high)>>1;
	
	int left_sum,right_sum,cross_sum;

	left_sum = maxSub(A, low, mid);
	
	right_sum= maxSub(A, mid+1, high);
	
	cross_sum= maxCenterSum(A, low, mid, high);

	/*return max(left_sum, max(right_sum, cross_sum));*/
	if(left_sum>=cross_sum && left_sum>=right_sum){flg=1;left_low=low;left_high=mid;return left_sum;}
	else if(right_sum>=left_sum && right_sum>=cross_sum){flg=3;right_low=mid+1;right_high=high; return right_sum;}
	else {flg=2;return cross_sum;}
}

int main()
{
	int T,i,n,cnt;
	int ans,a[100001],l,r;
	cin>>T;
	cnt=0;
	while(cnt<T)
	{
		flg=0;
		left_low=0;left_high=0;
		right_low=0;right_high=0;
		cross_low=0;cross_high=0;
		cnt++;
		cin>>n;
		for(i=1;i<=n;i++)
			cin>>a[i];
		ans = maxSub(a, 1, n);
		cout<<"Case "<<cnt<<":"<<endl<<ans<<" ";
		if(flg==0)cout<<"1 1"<<endl;
		else if(flg==1)cout<<left_low<<" "<<left_high<<endl;
		else if(flg==2)cout<<cross_low<<" "<<cross_high<<endl;
		else cout<<right_low<<" "<<right_high<<endl;
		if(cnt!=T)cout<<flg<<endl;
	}
	system("pause");
	return 0;
}

　　

//这个是改正了的！
//分治思想
/*
测试用例：
9
13 1 2 3 -100 1 2 3 -100 1 2 1 1 1
5 -3 -2 -1 -2 -3 
正确结果：
Case 1： 6 9 13
—— -1 3 3
*/
#include<iostream>
#define MIN -1000000005
using namespace std;

struct set
{
	int sum,begin,end;
};

//int left_low,left_high,right_low,right_high,cross_low,cross_high,flg;

set max(set a, set b){return a.sum>b.sum ? a:b;}

//求过中点的最大和
set maxCenterSum(int *A, int low, int mid, int high)
{
	int left_sum=MIN,sum=0,right_sum=MIN;
	int i,j;
	set cent;
	for(i=mid;i>=low;i--)
	{
		sum=sum+A[i];
		if(sum>=left_sum){left_sum=sum;cent.begin=i;}
	}
	sum=0;
	for(j=mid+1;j<=high;j++)
	{
		sum=sum+A[j];
		if(sum>=right_sum){right_sum=sum;cent.end=j;}
	}
	cent.sum = left_sum+right_sum;
	return cent;
}

//分别左右最大
set maxSub(int *A, int low, int high)
{
	set lcr;

	if(low==high){lcr.sum = A[low];lcr.begin=low;lcr.end=low;return lcr;}
	int mid=(low+high)>>1;

	set left_sum,right_sum,cross_sum;

	left_sum.begin=low;left_sum.end=mid;
	left_sum = maxSub(A, low, mid);

	right_sum.begin=mid+1;right_sum.end=high;
	right_sum= maxSub(A, mid+1, high);

	cross_sum= maxCenterSum(A, low, mid, high);

	return max(left_sum, max(right_sum, cross_sum));
}

int main()
{
	int T,i,n,cnt;
	int a[100001],l,r;
	set ans;
	cin>>T;
	cnt=0;
	while(cnt<T)
	{
		cnt++;
		cin>>n;
		for(i=1;i<=n;i++)
			cin>>a[i];
		ans = maxSub(a, 1, n);
		cout<<"Case "<<cnt<<":"<<endl<<ans.sum<<" ";
		cout<<ans.begin<<" "<<ans.end<<endl;
		if(cnt!=T)cout<<endl;
	}
	/*system("pause");*/
	return 0;
}

　　体会：还是应该善用结构体，像递归这种题不可能将所有结果仅仅保存在同一个int或其他类型下，必须要利用结构体。

     这种求最大子数组的问题，不便于用分治算法做，而应该用动态规划。分治的时间复杂度为nlgn，而动规的仅为n。