HDU 3415 Max Sum of Max-K-sub-sequence（单调队列+最大连续子串和）

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8433    Accepted Submission(s): 3095

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 

 

题意：

队列组成的圆，求长度不大于k的连续子串最大和，输出和 左端点l 右端点r。

POINT：

首位相连只要在n后面在加k个数字就可以了。先搞一个sum数组来保存前缀和。

若以l作为左端点来找最大的答案：

ans[l]=max{sum[l+i-1]}-sum[l-1]  (0<i<=k).

max{sum[l+i-1]} (0<i<=k) 这个部分可以用单调队列优化。

从1到n遍历左端点就可以了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int N = 100100+4;
const int INF=0x3f3f3f3f;
struct node
{
    int x,y;
}a[N<<1];
int sum[N<<1];
int n,k;
int ansmax,ansl,ansr;
deque<int> q;
void work()
{
    for(int i=1;i<k;i++)
    {
        while(!q.empty()&&sum[i]>sum[q.back()]) q.pop_back();
        q.push_back(i);
    }
    for(int i=1;i<=n;i++)
    {
        while(!q.empty()&&sum[i+k-1]>sum[q.back()]) q.pop_back();
        while(!q.empty()&&q.front()<i) q.pop_front();
        q.push_back(i+k-1);
        if(ansmax<sum[q.front()]-sum[i-1])
        {
            ansmax=sum[q.front()]-sum[i-1];
            ansr=q.front();
            ansl=i;
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
       // init();
        ansmax=-INF;
        scanf("%d %d",&n,&k);
        while(!q.empty()) q.pop_back();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].x);
            a[i].y=i;
            sum[i]=sum[i-1]+a[i].x;
        }
        for(int i=n+1;i<=n+k;i++)
        {
            a[i]=a[i-n];
            a[i].y=i;
            sum[i]=sum[i-1]+a[i].x;
        }
        work();
        if(ansr>n) ansr-=n;
        printf("%d %d %d\n",ansmax,ansl,ansr);
        
    }
}