HDu 3415 Max Sum of Max-K-sub-sequence単調キューdp
3096 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): 4581 Accepted Submission(s): 1656
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
Sample Output
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単調キュー最適化dpの入門問題
-----------------------
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4581 Accepted Submission(s): 1656
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
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単調キュー最適化dpの入門問題
-----------------------
#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[i]);
a[i+n]=a[i];
}
for (int i=1;i<=n+k;i++)
{
sum[i]+=sum[i-1]+a[i];
}
//f[i]=max(sum[i]-sum[k]);
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[i]=sum[i]-que[head];
if (f[i]>max_sum)
{
max_sum=f[i];
start=pt[head]+1;
end=i;
}
}
if (start>n) start=start-n;
if (end>n) end=end-n;
printf("%d %d %d
",max_sum,start,end);
}
return 0;
}