hdu 3415 Max Sum of Max-K-sub-sequence（单调队列）

最新推荐文章于 2023-04-13 09:00:14 发布

KinFungYu

最新推荐文章于 2023-04-13 09:00:14 发布

阅读量743

点赞数

CC 4.0 BY-SA版权

分类专栏： hdu 单调队列文章标签：单调队列

本文链接：https://blog.youkuaiyun.com/WEYuLi/article/details/20927121

hdu 同时被 2 个专栏收录

170 篇文章

订阅专栏

单调队列

2 篇文章

订阅专栏

探讨了如何解决环状序列中寻找长度不超过K的最大连续子序列和的问题，并提供了一种利用单调队列来高效计算的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5385 Accepted Submission(s): 1952

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

题意：给出一个长度为n的环状序列，求序列中长度不超过k的和最大的连续子序列
思路：环状序列看成长度为2*n的序列，然后求前缀和，求出max(sum[i] - sum[j])(i-k<j<i)。求这个最大值可以维护一个单调递增队列，队头元素始终为长度不超过k时最小的sum[i]的下标i。

AC代码：
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <algorithm>
#define ll long long
#define L(rt) (rt<<1)
#define R(rt)  (rt<<1|1)
#define eps 1e-6;
using namespace std;

const int INF = 1e9;
const int maxn = 100005;

int n, k, head, tail, s, t;
int num[maxn * 2], sum[maxn * 2];
int que[maxn * 2];
int main()
{
   int tt;
   scanf("%d", &tt);
   while(tt--)
   {
       scanf("%d%d", &n, &k);
       sum[0] = 0;
       for(int i = 1; i <= n; i++)
       {
           scanf("%d", &num[i]);
           sum[i] = num[i] + sum[i - 1];
       }
       for(int i = n + 1; i <= 2 * n; i++)
       sum[i] = num[i - n] + sum[i - 1];
       int ans = -INF;
       head = tail = 0;
       que[tail++] = 0;
       for(int i = 1; i <= 2 * n; i++)
       {
           while(head < tail && que[head] < i - k) head++;
           if(ans < sum[i] - sum[que[head]])
           {
               ans = sum[i] - sum[que[head]];
               s = que[head] + 1, t = i;
           }
           while(head < tail && sum[que[tail - 1]] > sum[i]) tail--;
           que[tail++] = i;
       }
       if(s > n) s -= n;
       if(t > n) t -= n;
       printf("%d %d %d\n", ans, s, t);
   }
   return 0;
}