hdu 3415 Max Sum of Max-K-sub-sequence

题目描述：

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4208    Accepted Submission(s): 1503

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

思路：
估计是dp
状态转移： dp[i]=sum[i]-min(sum[j])     (i-k<j<i)
维护sum[i] 最小就行
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define inf 0xfffffff
#define BUG puts("Here");system("pause")

using namespace std;
int const nMax =200010;
typedef long long LL;

LL dp[nMax];
LL sum[nMax];
int l[nMax];
int r[nMax];
LL a[nMax];
int n,k;

int q[nMax],he,ta;
void queue(int i){
    while(he<ta){
        int j=q[he],jj=q[he+1];
        if(i-j>k||sum[jj]<sum[j])he++;
        else break;
    }
    dp[i]=sum[i]-sum[q[he]];
    l[i]=q[he]+1;
    return ;
}
void insert(int i){
    while(he<ta){
        int j=q[ta];
        if(sum[i]<sum[j])ta--;
        else break;
    }
    q[++ta]=i;
    return ;
}


void DP()
{
    for(int i=1;i<=n+k;i++){
        queue(i);
        insert(i);
     //   printf("dp[%d]=%d\n",i,dp[i]);
    }
    LL ans=-inf; //printf("ans=%d\n",ans);
    int al=0,ar(0);
    for(int i=1;i<=n+k;i++)if(dp[i]>ans){

        ans=dp[i];// printf("  --dp[%d]=%d ans=%d\n",i,dp[i],ans);
        al=l[i];
        ar=i;
    }else if(ans==dp[i]){
        if(l[i]<al){
            ar=i;
            al=l[i];
        }else
        if(l[i]==al&&(ar-al>i-l[i])){
            ar=i;
           al=l[i];
       }
    }
    if(al>n)al-=n;
    if(ar>n)ar-=n;
    printf("%I64d %d %d\n",ans,al,ar);
    return ;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        sum[0]=0;
        dp[0]=0;
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        for(int i=n+1;i<=2*n;i++){
            sum[i]=sum[i-1]+a[i-n];
        }
       // BUG;
        ta=he=0;
        q[0]=0;
        DP();
    }
    return 0;
}