[HDU 3415] Max Sum of Max-K-sub-sequence

- 日理万妓 2021-12-03 07:17 223阅读 0赞

[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**

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 6 2

\-1 1 1

单调队列，子段和\[i, j\] = 前缀和\[1, i\] - 前缀和\[1, j-1\]。

单调队列可以解决定长 k 的单调序列区间中的最优解问题！

即： dp(i) = max\{ s(i-k+1), s(i-k+2),..., s(i) \}

或者： dp(i) = min\{ s(i-k+1), s(i-k+2),..., s(i) \}

为保证单调性和定长k的限制，在每次插入后，都要丢弃队列中超出k长度区域的值（即不在区间 i-k+1 ~ i 内的值），直到队首在给定的k长度范围内，同时队首元素就是当前最优解。

1 #include <vector>
     2 #include <cstdio>
     3 #include <queue>
     4 #include <algorithm>
     5 #include <climits>
     6 using namespace std;
     7 
     8 int T;
     9 int N, K;
    10 vector<int> v;
    11 deque<int> q;
    12 
    13 void solve() {
    14     int sum = INT_MIN, start = 0, end = 0;
    15     for (int i = 1; i <= N + K; ++i) {
    16         while (!q.empty() && v[q.back()] > v[i]) q.pop_back();
    17         q.push_back(i - 1);
    18         while (i - q.front() > K) q.pop_front();
    19         if (sum < v[i] - v[q.front()]) {
    20             sum = v[i] - v[q.front()];
    21             start = q.front() + 1;
    22             end = i;
    23         }
    24     }
    25     printf("%d %d %d\n", sum, start, end > N ? end - N : end);
    26 }
    27 
    28 int main() {
    29     scanf("%d", &T);
    30     while (T--) {
    31         scanf("%d %d", &N, &K);
    32         v.resize(N + K + 1, 0);
    33         for (int i = 1; i <= N; ++i) {
    34             scanf("%d", &v[i]);
    35             v[i] += v[i-1];
    36         }
    37         for (int i = N + 1; i <= N + K; ++i) {
    38             v[i] = v[N] + v[i-N];
    39         }
    40         q.clear();
    41         solve();
    42     }
    43     return 0;
    44 }

转载于:https://www.cnblogs.com/easonliu/p/4582694.html

[Max Sum of Max-K-sub-sequence]: http://acm.hdu.edu.cn/showproblem.php?pid=3415