HDU 3415——Max Sum of Max-K-sub-sequence

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

 

 

Author

shǎ崽@HDU

 

 

Source

HDOJ Monthly Contest – 2010.06.05

 

 

Recommend

lcy

 

【算法分析】

 

和 1003 Max Sum 差不多，不同之处有两点；

1、数组是循环的

2、限制了最长的连续字段的长度为 K

还是最原始的想法：找最大的 s[i] - s[j] 只不过 i - K <= j < i 了

那么还是可以用单调队列优化，为什么？

区间：[i - K , i] 是单调的

接着的东西就跟 1003 差不多了

PS：由于数组定义小了，硬是WA了将近一个小时！以后一定要注意数据的大小！！！

下面的代码仅供参考：

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define INF 0x7fffffffusing namespace std;int main(){    int T,N,K,st,en,Max;    int a[200005],q[200005],head,tail;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&K);        a[0]=0;        for(int i=1;i<=N;i++)        {            scanf("%d",&a[i]);            a[i]+=a[i-1];        }        for(int i=1;i<=K;i++)            a[N+i]=a[N]+a[i];        head=tail=0;        q[++tail]=0;        Max=-INF;        for(int i=1;i<=N+K;i++)        {            while(head<tail && q[head+1]<i-K) head++;            if(a[i]-a[q[head+1]]>Max)            {                Max=a[i]-a[q[head+1]];                st=q[head+1]+1;                en=i;            }            while(head<tail && a[q[tail]]>a[i]) tail--;            q[++tail]=i;        }        st=(st-1)%N+1;        en=(en-1)%N+1;        printf("%d %d %d\n",Max,st,en);    }    return 0;}