HDOJ 3415 Max Sum of Max-K-sub-sequence

单调队列优化。。。。

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4759    Accepted Submission(s): 1734

Problem Description

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output a 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 minimum start position, if still more than one , output the minimum length of them.

Sample Input

4
6 3
6 -1 2 -6 5 -5
6 4
6 -1 2 -6 5 -5
6 3
-1 2 -6 5 -5 6
6 6
-1 -1 -1 -1 -1 -1

Sample Output

7 1 3
7 1 3
7 6 2
-1 1 1

Author

shǎ崽@HDU

Source

HDOJ Monthly Contest – 2010.06.05

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#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a[411111];

int f[411111];

int que[1111111];

int pt[1111111];

int n,k;

int T;

int head,tail;

int sum[411111];

int max_sum,start,end;

int main()

{

scanf("%d",&T);

while (T--)

{

memset(f,0,sizeof(f));

memset(que,0,sizeof(que));

memset(pt,0,sizeof(pt));

memset(sum,0,sizeof(sum));

scanf("%d%d",&n,&k);

for (int i=1;i<=n;i++)

{

scanf("%d",&a);

a[i+n]=a;

}

for (int i=1;i<=n+k;i++)

{

sum+=sum[i-1]+a;

}

head=tail=0;

max_sum=start=end=-1e9;

for(int i=1;i<=n+k;i++)

        {

            while(head<tail&&i-pt[head]>k)  head++;

            while(head<tail&&sum[i-1]<que[tail-1]) tail--;

            que[tail]=sum[i-1]; pt[tail++]=i-1;

            f=sum-que[head];

            if(max_sum<f)

            {

                max_sum=f;

                start=pt[head]+1;

                end=i;

            }

        }

if(start>n) start-=n;

if(end>n) end-=n;

printf("%d %d %d\n",max_sum,start,end);

}

return 0;

}