HDU 3415(单调队列)

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4080 Accepted Submission(s): 1453

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

/*
题目大意：给出一个有N个数字（N<=10^5）的环状序列，让你求一个和最大的连续子序列。这个连续子序列的长度小于等于K。
输出和，起始和结束位置 
*/
#include<iostream>
#include<queue>
using namespace std;

const int INF = 0x3fffffff;
const int maxn = 100010;
int num[maxn],sum[maxn];

int main()
{
    int T,i,j;
    int N,K,n;
    cin>>T;
    while(T--)
    {
        cin>>N>>K;
        sum[0]=0;
        for(i=1;i<=N;i++)
        {
            cin>>num[i];
            sum[i]=sum[i-1]+num[i];
        }
        for(i=N+1;i<N+K;i++)
        {
            sum[i]=sum[i-1]+num[i-N];
        }
        n=N+K-1; 
        deque <int> q;
        q.clear();
        int ans=-INF;
        int start,end;
        //枚举以j结尾的区间
        for(j=1;j<=n;j++)
        {
            while(!q.empty() && sum[j-1]<sum[q.back()])
                q.pop_back();
            while(!q.empty() && q.front()<(j-K))//区间最大长度是K ,没有等号，因为入队的是 (j-1)
                q.pop_front();
            q.push_back(j-1);
            if(sum[j]-sum[q.front()]>ans)
            {
                ans=sum[j]-sum[q.front()];
       //i到j和是,sum[8] - sum[5] = sum[6]+sum[7]+sum[8]



　　　　　　         start=q.front()+1;
                end=j;
            }
        }
        cout<<ans<<" "<<start<<" "<<(end>N?end%N:end)<<endl;
    }
    return 0;
}
////暴力枚举起点和终点估计会超时