單調隊列優化。。。。

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4759????Accepted Submission(s): 1734


Problem DescriptionGiven 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.
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 two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).
OutputFor 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 Input4
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 Output7 1 3
7 1 3
7 6 2
-1 1 1
Authorshǎ崽@HDU
SourceHDOJ Monthly Contest – 2010.06.05
Recommendlcy

#include <iostream>#include <cstdio>#include <cstring>
using namespace std;
int a[411111];int f[411111];int que[1111111];int pt[1111111];int n,k;int T;int head,tail;int sum[411111];int max_sum,start,end;

int main(){scanf("%d",&T);while (T--){memset(f,0,sizeof(f));memset(que,0,sizeof(que));memset(pt,0,sizeof(pt));memset(sum,0,sizeof(sum));scanf("%d%d",&n,&k);for (int i=1;i<=n;i++){scanf("%d",&a);a[i+n]=a;}for (int i=1;i<=n+k;i++){sum+=sum[i-1]+a;}
head=tail=0;max_sum=start=end=-1e9; for(int i=1;i<=n+k;i++)? ? ? ? {? ? ? ? ? ? while(head<tail&&i-pt[head]>k) ?head++;? ? ? ? ? ? while(head<tail&&sum[i-1]<que[tail-1]) tail--;
? ? ? ? ? ? que[tail]=sum[i-1]; pt[tail++]=i-1;
? ? ? ? ? ? f=sum-que[head];
? ? ? ? ? ? if(max_sum<f)? ? ? ? ? ? {? ? ? ? ? ? ? ? max_sum=f;? ? ? ? ? ? ? ? start=pt[head]+1;? ? ? ? ? ? ? ? end=i;? ? ? ? ? ? }? ? ? ? }
if(start>n) start-=n;if(end>n) end-=n;
printf("%d %d %d\n",max_sum,start,end);
}return 0;}

轉載于:https://www.cnblogs.com/CKboss/p/3350984.html

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