首页 > 代码库 > hdu 3415 Max Sum of Max-K-sub-sequence

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

2024-07-07 01:12:37   228人阅读

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

 单调队列的应用

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;
const int maxn=100000+10;
int sum[maxn*2],a[maxn];
int main()
{
    int n,k,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        sum[0]=0;
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        for (int i=1;i<=n;i++)
        sum[n+i]=sum[n+i-1]+a[i];
        int start,end,ans=-99999999;
        deque<int>q;
        //q.push_back(0);
        for (int i=1;i<=n+k-1;i++)
        {
            while(!q.empty() && sum[i-1]<sum[q.back()])
            q.pop_back();
            while(!q.empty() && q.front()<i-k)
            q.pop_front();
            q.push_back(i-1);
            if (sum[i]-sum[q.front()]>ans)
            {
                ans=sum[i]-sum[q.front()];
                start=q.front()+1;
                end=i;
            }
        }
        if (end!=n) end%=n;
        printf("%d %d %d\n",ans,start,end);
    }
    return 0;
}

作者 chensunrise


联系
我们