hdu3415(单调队列）
发布时间：2020-12-15 07:59:49 所属栏目：Java 来源：网络整理
导读：Max Sum of Max-K-sub-sequence Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10481????Accepted Submission(s): 3853 Problem Description Given a circle sequence A[1],A[2],A[3]......A[
Max Sum of Max-K-sub-sequenceTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
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.
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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).
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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.
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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
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Sample Output
7 1 3 7 1 3 7 6 2 -1 1 1
用sum数组存前缀和，a【i.....j]=sum[j]-sum[i-1];max(a[1....j])=sum[j]-min(sum[i-1])? ? j-(i-1)<=k;
用单调栈维护最小的sum[i-1]即可；
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;int a[100005],sum[200100],q[200100];
int start,endi,total;int res;int ans;
void get()
{
int head=0,tail=0;
ans=-100000000;
for(int i=1;i<total;i++)
{
while(head<tail&&sum[q[tail-1]]>=sum[i-1])tail--;
while(head<tail&&q[head]<i-m)head++;
q[tail++]=i-1;
if(sum[i]-sum[q[head]]>ans)
{
start=q[head];
endi=i;
ans=sum[endi]-sum[start];
}
}
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
cin>>n>>m;sum[0]=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
sum[i]=sum[i-1]+a[i];
}
for(int i=1;i<m;i++)
sum[n+i]=sum[n+i-1]+a[i];
total=n+m;
get();
start++;
if(start>n)start-=n;
if(endi>n)endi-=n;
cout<<ans<<" "<<start<<" "<<endi<<endl;
}
return 0;
}
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