HDU 3415(単調キュー)
8468 ワード
Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4080 Accepted Submission(s): 1453
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
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4080 Accepted Submission(s): 1453
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
1 /*
2 ： N （N<=10^5） ， 。 K。
3 ，
4 */
5 #include<iostream>
6 #include<queue>
7 using namespace std;
8
9 const int INF = 0x3fffffff;
10 const int maxn = 100010;
11 int num[maxn],sum[maxn];
12
13 int main()
14 {
15 int T,i,j;
16 int N,K,n;
17 cin>>T;
18 while(T--)
19 {
20 cin>>N>>K;
21 sum[0]=0;
22 for(i=1;i<=N;i++)
23 {
24 cin>>num[i];
25 sum[i]=sum[i-1]+num[i];
26 }
27 for(i=N+1;i<N+K;i++)
28 {
29 sum[i]=sum[i-1]+num[i-N];
30 }
31 n=N+K-1;
32 deque <int> q;
33 q.clear();
34 int ans=-INF;
35 int start,end;
36 // j
37 for(j=1;j<=n;j++)
38 {
39 while(!q.empty() && sum[j-1]<sum[q.back()])
40 q.pop_back();
41 while(!q.empty() && q.front()<(j-K))// K , ， (j-1)
42 q.pop_front();
43 q.push_back(j-1);
44 if(sum[j]-sum[q.front()]>ans)
45 {
46 ans=sum[j]-sum[q.front()];
47 //i j ,sum[8] - sum[5] = sum[6]+sum[7]+sum[8]
start=q.front()+1;
48 end=j;
49 }
50 }
51 cout<<ans<<" "<<start<<" "<<(end>N?end%N:end)<<endl;
52 }
53 return 0;
54 }
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