1.题目描述：

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8024    Accepted Submission(s): 2929 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   Author shǎ崽@HDU   Source HDOJ Monthly Contest – 2010.06.05   Recommend lcy   |   We have carefully selected several similar problems for you:   3423  3417  3418  3419  3421 

2.题意概述：

要你寻找一个连续子序列，使得他们的和最大，且序列长度不超过K

3.解题思路：

因为序列是环状的，所以可以在序列后面复制前k-1个数字。如果用s[i]来表示复制过后的序列的前i个数的和，那么任意一个子序列[i..j]的和就等于s[j]-s[i-1]。 对于每一个j，用s[j]减去最小的一个s[i](i>=j-k)就可以得到以j为终点长度不大于k的和最大的序列了。将原问题转化为这样一个问题后，就可以用单调队列解决了。 单调队列即保持队列中的元素单调递增（或递减）的这样一个队列，可以从两头删除，只能从队尾插入。单调队列的具体作用在于，由于保持队列中的元素满足单调性， 对于上述问题中的每个j，可以用O(1)的时间找到对应的s[i]。（保持队列中的元素单调递增的话，队首元素便是所要的元素了）。 维护方法：对于每个j，我们插入s[j-1]的下标，插入时从队尾插入。为了保证队列的单调性，我们从队尾开始删除元素，直到队尾元素对应的值比当前需要插入的s[j-1]小， 就将当前元素下标插入到队尾。之所以可以将之前的队列尾部元素全部删除，是因为它们已经不可能成为最优的元素了，因为当前要插入的元素位置比它们靠前， 对应的值比它们小。我们要找的，是满足(i>=j-k)的i中最小的s[i]。在插入元素后，从队首开始，将不符合限制条件(i<j-k)的元素全部删除，此时队列一定不为空。（因为刚刚插入了一个一定符合条件的元素）。  4.AC代码：

#include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <functional> #include <cmath> #include <vector> #include <queue> #include <deque> #include <stack> #include <map> #include <set> #include <ctime> #define INF 0x3f3f3f3f #define maxn 200100 #define lson root << 1 #define rson root << 1 | 1 #define lent (t[root].r - t[root].l + 1) #define lenl (t[lson].r - t[lson].l + 1) #define lenr (t[rson].r - t[rson].l + 1) #define N 1111 #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) using namespace std; const int mod = 1e9 + 7; typedef long long ll; int q[maxn], a[maxn], sum[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif int t, n, k; scanf("%d", &t); while (t--) { scanf("%d%d", &n, &k); sum[0] = 0; for (int i = 1; i <= n; i++) { scanf("%d", &a[i]); sum[i] = sum[i - 1] + a[i]; } for (int i = n + 1; i <= n * 2; i++) sum[i] = sum[i - 1] + a[i - n]; int ans = -INF, l = -1, r = -1, head = 0, tail = 0; q[0] = 0; for (int i = 1; i <= n + k; i++) { while (head <= tail && sum[q[tail]] > sum[i - 1]) tail--; q[++tail] = i - 1; while (i - q[head] > k) head++; int tmp = sum[i] - sum[q[head]]; if (ans < tmp) { ans = tmp; l = q[head] + 1; r = i; } } printf("%d %d %d\n", ans, l > n ? l - n : l, r > n ? r - n : r); } #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; }