hdu Max Sum of Max-K-sub-sequence（单调队列）

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

46 36 -1 2 -6 5 -56 46 -1 2 -6 5 -56 3-1 2 -6 5 -5 66 6-1 -1 -1 -1 -1 -1

Sample Output

7 1 37 1 37 6 2-1 1 1

题意：给定n个数组成的循环序列，求最大的连续子段和

单调队列简单题，由于原序列是一个环，在序列后面复制一段再处理会更方便一点

代码：
#include<stdio.h>int a[200010],sum[200010];   // a记录队列中元素位置，sum记录前i项和int l,r,t,n,k,p,q,max;int main(){    scanf("%d",&t);    while (t--)    {        scanf("%d%d",&n,&k);        for (int i=1;i<=n;i++)        {            scanf("%d",&sum[i]);            sum[i]+=sum[i-1];        }        for (int i=n+1;i<=n+k;i++)            sum[i]=sum[n]+sum[i-n];        l=0;r=-1;max=-2000000000;        for (int i=1;i<=n+k;i++)        {            while (l<=r&&sum[a[r]-1]>sum[i-1])                      r--;               a[++r]=i;            while (i-a[l]>=k)                     l++;             if (sum[i]-sum[a[l]-1]>max)            {                max=sum[i]-sum[a[l]-1];                p=a[l];                q=i;            }        }        printf("%d %d %d\n",max,p,q>n?q-n:q);    }    return 0;}