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

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8462    Accepted Submission(s): 3111

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

 

题目链接：HDU 3415

把原数组复制一接到末尾，完成了从环到链的转换，然后用差分的思想，用$presum[r]-presum[l-1]$表示区间的和，那只要对于每一个$presum[r]$，找到在他前面的连续k个$presum[j]$中的最小值即可，其中用单调队列可以优化到O(N)的复杂度

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 200010;
int arr[N], pre[N];

int main(void)
{
    int T, n, k, i;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &k);
        for (i = 1; i <= n; ++i)
        {
            scanf("%d", &arr[i]);
            arr[n + i] = arr[i];
        }
        int nn = n << 1;
        for (i = 1; i <= nn; ++i)
            pre[i] = pre[i - 1] + arr[i];
        deque<int>q;
        int ans = -1e9 - 7, l = 1, r = 1;
        for (i = 1; i <= n + k - 1; ++i)
        {
            while (!q.empty() && pre[q.back()] > pre[i - 1])
                q.pop_back();
            q.push_back(i - 1);
            while (!q.empty() && i - q.front() > k)
                q.pop_front();
            int v = pre[i] - pre[q.front()];
            if (v > ans)
            {
                ans = v;
                l = q.front() + 1;
                r = i;
            }
            else if (v == ans && q.front() + 1 < l)
            {
                l = q.front() + 1;
                r = i;
            }
            else if (v == ans && q.front() + 1 == l && i - q.front() < r - l + 1)
                r = i;
        }
        l %= n;
        r %= n;
        if (!l)
            l = n;
        if (!r)
            r = n;
        printf("%d %d %d\n", ans, l, r);
    }
    return 0;
}