单调队列的一个应用——求解连续区间最大值（HDU Max Sum of Max-K-sub-sequence）

Description

Given a circle sequence A[1],A[2],A[3]......A
. Circle sequence means the left neighbour of A[1] is A
, and the right neighbour of A
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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 100010;
int N, K;
int a[MAXN];
int sum[2 * MAXN];

/*
* 由于是统计不超过K个连续元素和的最大值，我们以sum值作为处理对象。
* 思想：线性枚举当前状态，并且维护在K范围内的最小值，对就是用单调
* 队列取维护当前允许范围内的最小值，然后求出当前最有解去更新总体的最优解即可
* 很简单的，实际上就应该是传说中的斜率优化。理解万岁，为什么当年就是想不出来呢？
*/

struct node {
int num, index;
} que[2 * MAXN];
int front, tail;

void init() {
front = 0, tail = -1;
}

void Insert(int pos, int num) {
if (front > tail) {
front = tail = 0;
que[front].index = pos;
que[front].num = num;
return;
}
while (front <= tail && que[tail].num >= num) tail--;
que[++tail].index = pos;
que[tail].num = num;
while (front <= tail && pos - que[front].index + 1 > K) front++;
}

int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d", &N, &K);
init();
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[i + N] = sum[i + N - 1] + a[i];
}
int l, r;
int ans = -100000000;
Insert(0, sum[0]);
for (int i = 1; i <= 2 * N; i++) {
if (sum[i] - que[front].num > ans) {
ans = sum[i] - que[front].num;
l = que[front].index + 1;
r = i;
}
Insert(i, sum[i]);
}
if (l > N) l -= N;
if (r > N) r -= N;
printf("%d %d %d\n", ans, l, r);
}
return 0;
}