hdu 3415 Max Sum of Max-K-sub-sequence(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): 3728    Accepted Submission(s): 1317

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

 

Recommend

lcy

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3415

题意：给你一个环形的数列，问你连续长度为k的数的和的最大值，并且输出区间坐标，使坐标字典序最小

分析：把环形转换成直线处理即可，求出累加和数组，用优先队列维护

PS：这回总是是1Y了，就是这种感觉^_^

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=222222;
int a[mm],q[mm];
int main()
{
    int i,l,r,n,k,t,tmp,ans,u,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        ans=-2e9;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&a[i]);
            a[i+n]=a[i];
            if(a[i]>ans)
                ans=a[i],u=i-1,v=i;
        }
        a[0]=0;
        for(i=1;i<n*2;++i)
            a[i]+=a[i-1];
        q[l=r=0]=0;
        for(i=1;i<n*2;++i)
        {
            while(l<=r&&a[q[r]]>a[i])--r;
            q[++r]=i;
            while(i-q[l]>k)++l;
            if(q[l]<i)
            {
                tmp=a[i]-a[q[l]];
                if(tmp==ans&&q[l]+1<u)
                    u=q[l],v=i;
                if(tmp>ans)
                {
                    ans=tmp;
                    u=q[l],v=i;
                }
            }
        }
        printf("%d %d %d\n",ans,(u%n)+1,((v-1)%n)+1);
    }
    return 0;
}