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HDU 3415 Max Sum of Max-K-sub-sequence
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Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5791 Accepted Submission(s): 2083
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
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
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](为什么不是s[j]? 队列里面维护的是区间开始的下标,j是区间结束的下标),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,直到队尾元素比当前需要插入的元素优(本题中是值比待插入元素小,位置比待插入元素靠前,不过后面这一个条件可以不考虑),就将当前元素插入到队尾。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,值比它们小。我们要找的,是满足(i>=j-k+1)的i中最小的s[i],位置越大越可能成为后面的j的最优s[i]。
在插入元素后,从队首开始,将不符合限制条件(i>=j-k+1)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)
代码如下:(看了别人的才做出来的,汗)
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> #include <iostream> using namespace std; #define INF 0x3fffffff int sum[100047],a[200047]; int main() { int t,i,n,m,k,head,end; scanf("%d",&t); while(t--) { memset(sum,0,sizeof(sum)); scanf("%d%d",&n,&k); for(i = 1 ; i <= n ; i++) { scanf("%d",&a[i]); sum[i] = sum[i-1]+a[i];//将前i项和全部存入sum数组中 } for(i = n+1 ; i < n+k ; i++) { sum[i] = sum[i-1]+a[i-n];//将前n+k-1项和全部存入sum数组中 } int ans = -INF;//初始化ans为最小 deque<int>Q; Q.clear();//清空双向队列 for(i = 1 ; i < n+k ; i++) { while(!Q.empty() && sum[i-1] < sum[Q.back()])//保持队列的单调性(递增) Q.pop_back(); while(!Q.empty() && i-k > Q.front())//超过k的长度则消除队列前面的元素 Q.pop_front(); Q.push_back(i-1); if(ans < sum[i]-sum[Q.front()])//如果当前的值比ans大就更新ans的值 { //记录,sum[n]-sum[m]所得出的是n-1到m+1之间的和 ans = sum[i]-sum[Q.front()]; head = Q.front()+1; end = i; } } if(end > n)//标记的点大于了n则循环 end%=n; if(head > n)//标记的点大于了n则循环 head%=n; printf("%d %d %d\n",ans,head,end); } return 0; }
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