HDU3415
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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.
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).
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
Author
shǎ崽@HDU
Source
HDOJ Monthly Contest – 2010.06.05
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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)。
#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;}
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