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。