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

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题意：求长度不超过K的最大的连续序列的和

思路：采用单调队列，我们要求的是Max{sum[i]-sum[j]}(i-j<=k),可以这么想每次的i，我们都在可以的范围内找个一个最小的sum[j]就是可以了，最后求最大就是了，至于怎么能够快速的找个一个最小的数，我们采用单调队列

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1000005;const int INF = 0x3f3f3f3f;int n,k;int arr[MAXN],sum[MAXN],q[MAXN];int main(){int t;scanf("%d", &t);while (t--){scanf("%d%d", &n, &k);for (int i = 1; i <= n; i++){scanf("%d", &arr[i]);arr[i+n] = arr[i];}sum[0] = 0;for (int i = 1; i <= 2*n; i++)sum[i] = sum[i-1] + arr[i];int head = 0,tail = 0;q[head] = 0;int Max = -INF,x,y;for (int i = 1; i <= 2*n; i++){while (head <= tail && i-q[head]>k)head++;int j = q[head];if (sum[i] - sum[j] > Max){Max = sum[i] - sum[j];x = j+1;y = i;}while (head <= tail && sum[q[tail]] > sum[i])tail--;q[++tail] = i;}if (x > n)x -= n;if (y > n)y -= n;printf("%d %d %d\n", Max, x, y);}return 0;}

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