HDU-3415-Max Sum of Max-K-sub-sequence(单调队列,带限制的最大子段和)

链接:

https://vjudge.net/problem/HDU-3415

题意:

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.

思路:

单调队列,维护一个上升的前缀和,最大值就是Sum[i]-Sum[j],而取不到位置从前面pop,比当前位置大的值从后面pop.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 1e5+10;
const int INF = 1e9;

int a[MAXN*2];
int Sum[MAXN*2];
int n, k;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        memset(Sum, 0, sizeof(Sum));
        cin >> n >> k;
        for (int i = 1; i <= n; i++)
            cin >> a[i], a[i + n] = a[i];
        for (int i = 1; i <= n*2; i++)
            Sum[i] = Sum[i - 1] + a[i];
        deque<pair<int, int> > que;
        int res = Sum[1], l=0, r=1;
        for (int i = 0; i <= 2 * n; i++)
        {
            while (!que.empty() && i-que.front().first > k)
                que.pop_front();
            while (!que.empty() && Sum[i] < que.back().second)
                que.pop_back();
            if (!que.empty() && Sum[i]-que.front().second > res)
            {
                res = Sum[i]-que.front().second;
                l = que.front().first;
                r = i;
//                cout << l << ' ' << r << endl;
            }
            que.push_back(make_pair(i, Sum[i]));
        }
        cout << res << ' ' << ((l+1) > n ? l+1-n:l+1) << ' ' << (r > n?r-n:r) << endl;
    }

    return 0;
}