hdu3415 Max Sum of Max-K-sub-sequence

 

 Max Sum of Max-K-sub-sequence

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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
        
        
        
        

 

      
      
      
      

       
       
       
       题目优先队列
      
      
      
      
      
      
      
      

       
       
       
       ，在1-n加一个n-1就可以把环转化成一条线，用sum求合，那么从i到j的和就可以用sum[j]-sum[i],这个技巧也可以优化求和！然后把sum[i]用单调队列，j从0到n+m；这样一个一个求和，就可以了！这里的，最大值，用优先队列保存，这样就可以把复杂度压缩在nlgn了
      
      
      
      
      
      
      
      

       
       
       
       
#include <iostream>
#include<stdio.h>
using namespace std;
int num[250000],sum[250000],prim[250000];
int main()
{
    int n,m,t,i,front,rear;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        sum[0]=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            sum[i]=sum[i-1]+num[i];//用合来简化运算
        }
        for(;i<=2*n;i++)
        {
           sum[i]=sum[i-1]+num[i-n];//大于Ｎ的部分i-n对应的相应的ＮＵＭ

        }
        front=0;
        rear=0;
        int maxx=-1e10,sx=0,ex=0;
        for(i=1;i<=n+m;i++)
        {
            while(front<rear&&sum[prim[rear-1]]>sum[i-1])//插入
            {
                rear--;
            }

            prim[rear++]=i-1;
            while(front<rear&&i-prim[front]>m)//去掉过界的
            {
                front++;
            }
            if(maxx<sum[i]-sum[prim[front]])//保存最大值，和相应的坐标
            {
                sx=prim[front]+1;
                ex=i;
                maxx=sum[i]-sum[prim[front]];
            }

        }
        if(sx>n)sx-=n;//注意大于n的其实是构造的模型，再重新
        if(ex>n)ex-=n;
        printf("%d %d %d\n",maxx,sx,ex);

    }
    return 0;
}