Monotonic Queue Optimization DP

Max Sum of Max-k-sub-sequence

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 6277 Accepted Submission (s): 2289

Problem Descriptiongiven a circle sequence a[1],a[2],a[3] ... A[n]. Circle sequence means the left neighbour of A[1] was a[n], and the right neighbour of A[n] is a[1].
Now your job was 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.

Inputthe 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 the integers n, K (1<=n<=100000, 1<=k<=n), then N integers followe D (all the integers is between-1000 and 1000).

Outputfor, should output a line contains three integers, the Max Sum in the sequence, the start Positio N of the sub-sequence, the end position of the sub-sequence. If there is more than one result, output the minimum start position, if still more than one, output the minimum length O F them.

Sample Input46 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 Output7 1 37 1 37 6 2-1 1 1

Authorshǎ Cub @hdu

Sourcehdoj Monthly contest–2010.06.05

#include <iostream> #include <cstring> #include <cstdio> #include <string> #include <cmath > #include <algorithm> #include <cstdlib> #include <cmath> #include <queue> #include < Vector> #include <set>using namespace std; #define INF 100000000int Tt,n,k;int sum[200005],a[200005],hd,ed;int    Main () {scanf ("%d", &AMP;TT);        while (tt--) {scanf ("%d%d", &n,&k);        int j=n;        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+k;i++) {sum[i]=sum[i-1]+a[i-n];        } n=n+k-1;        int ans=-inf;        Deque<int> Q;        Q.clear (); for (int i=1;i<=n;i++) {while (!q.empty () &&sum[i-1]<sum[q.back ()))//Maintain monotonicity Q.            Pop_back ();            while (!q.empty () &&q.front () <i-k) Q.pop_front ();       Q.push_back (i-1);     if (Sum[i]-sum[q.front ()]>ans) {Ans=sum[i]-sum[q.front ()];                Hd=q.front (+1);            Ed=i;        }} if (Ed>j) ed=ed%j;    printf ("%d%d%d\n", ans,hd,ed); } return 0;}

　　

Monotonic Queue Optimization DP