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[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

46 36 -1 2 -6 5 -56 46 -1 2 -6 5 -56 3-1 2 -6 5 -5 66 6-1 -1 -1 -1 -1 -1

 

Sample Output

7 1 37 1 37 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;}